diff --git a/typeset/converter.js b/typeset/converter.js
index 74447c8..a9536b4 100644
--- a/typeset/converter.js
+++ b/typeset/converter.js
@@ -74,7 +74,7 @@ const injectMathJax = async (log, inputPath, cssPath, outputPath, mathJaxPath) =
})
log.debug(`Opened "${url}"`)
- await page.evaluate(() => {
+ await page.evaluate(/* istanbul ignore next */() => {
window.__TYPESET_CONFIG = {
isDone: false,
isFailed: false,
@@ -83,7 +83,7 @@ const injectMathJax = async (log, inputPath, cssPath, outputPath, mathJaxPath) =
})
log.debug(`Injecting CSS...`)
- await page.evaluate(stylePath => {
+ await page.evaluate(/* istanbul ignore next */stylePath => {
if (stylePath) {
console.log('Setting stylesheets...')
const style = document.createElement('link')
@@ -108,7 +108,7 @@ const injectMathJax = async (log, inputPath, cssPath, outputPath, mathJaxPath) =
// Typeset equations
log.info(`Injecting MathJax (and typesetting)...`)
- const didMathJaxLoad = await page.evaluate((mathJaxPath) => {
+ const didMathJaxLoad = await page.evaluate(/* istanbul ignore next */(mathJaxPath) => {
console.log('Setting config for MathJax...')
const MATHJAX_CONFIG = {
extensions: ['mml2jax.js', 'MatchWebFonts.js'],
@@ -187,7 +187,7 @@ const injectMathJax = async (log, inputPath, cssPath, outputPath, mathJaxPath) =
log.info(`Polling to see when MathJax is done typesetting...`)
let pageContentAfterSerialize = ''
while (true) {
- const {isFailed, isDone} = await page.evaluate(() => {
+ const {isFailed, isDone} = await page.evaluate(/* istanbul ignore next */() => {
if (!window.MathJax) {
console.error('MathJax was not loaded')
return {isFailed: true}
@@ -207,7 +207,7 @@ const injectMathJax = async (log, inputPath, cssPath, outputPath, mathJaxPath) =
return STATUS_CODE.ERROR
} else if (isDone) {
log.info('Serializing document...')
- pageContentAfterSerialize = await page.evaluate(() => {
+ pageContentAfterSerialize = await page.evaluate(/* istanbul ignore next */() => {
// Remove any elements we added
window.__TYPESET_CONFIG.elementsToRemove.forEach(el => el.remove())
diff --git a/typeset/tests/test-output.html b/typeset/tests/test-output.html
index fe0b69e..7b65a30 100644
--- a/typeset/tests/test-output.html
+++ b/typeset/tests/test-output.html
@@ -46,7 +46,16 @@
.MathJax .MathJax_HitBox {cursor: text; background: white; opacity: 0; filter: alpha(opacity=0)}
.MathJax .MathJax_HitBox * {filter: none; opacity: 1; background: transparent}
#MathJax_Tooltip * {filter: none; opacity: 1; background: transparent}
-@font-face {font-family: MathJax_Blank; src: url('about:blank')}
+@font-face {font-family: MathJax_Main; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Main-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Main-Regular.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Main-bold; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Main-Bold.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Main-Bold.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Main-italic; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Main-Italic.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Main-Italic.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Math-italic; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Math-Italic.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Math-Italic.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Caligraphic; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Caligraphic-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Caligraphic-Regular.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Size1; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Size1-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Size1-Regular.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Size2; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Size2-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Size2-Regular.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Size3; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Size3-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Size3-Regular.otf?V=2.7.4') format('opentype')}
+@font-face {font-family: MathJax_Size4; src: url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/woff/MathJax_Size4-Regular.woff?V=2.7.4') format('woff'), url('file:///C:/Users/Piotr/Documents/GitHub/bakedpdf/node_modules/mathjax/unpacked/../fonts/HTML-CSS/TeX/otf/MathJax_Size4-Regular.otf?V=2.7.4') format('opentype')}
+
@@ -511,21 +520,21 @@
In the absence of friction, the time to complete one oscillation remains constant and is called the period (T). Its units are usually seconds, but may be any convenient unit of time. The word ‘period’ refers to the time for some event whether repetitive or not, but in this chapter, we shall deal primarily in periodic motion, which is by definition repetitive.
A concept closely related to period is the frequency of an event. Frequency (f) is defined to be the number of events per unit time. For periodic motion, frequency is the number of oscillations per unit time. The relationship between frequency and period is
The SI unit for frequency is the hertz (Hz) and is defined as one cycle per second:
-
1Hz=1cyclesecor1Hz=1s=1s−1.
+
1Hz=1cyclesecor1Hz=1s=1s−1.
A cycle is one complete oscillation.
Example 15.1
Determining the Frequency of Medical Ultrasound
-Ultrasound machines are used by medical professionals to make images for examining internal organs of the body. An ultrasound machine emits high-frequency sound waves, which reflect off the organs, and a computer receives the waves, using them to create a picture. We can use the formulas presented in this module to determine the frequency, based on what we know about oscillations. Consider a medical imaging device that produces ultrasound by oscillating with a period of 0.400μs. What is the frequency of this oscillation?
+Ultrasound machines are used by medical professionals to make images for examining internal organs of the body. An ultrasound machine emits high-frequency sound waves, which reflect off the organs, and a computer receives the waves, using them to create a picture. We can use the formulas presented in this module to determine the frequency, based on what we know about oscillations. Consider a medical imaging device that produces ultrasound by oscillating with a period of
0.400μs. What is the frequency of this oscillation?
Strategy
The period (T) is given and we are asked to find frequency (f).
Solution
-Substitute 0.400μs for T in f=1T:
-
+Substitute
0.400μs for
T in
f=1T:
+
Solve to find
-
+
Significance
This frequency of sound is much higher than the highest frequency that humans can hear (the range of human hearing is 20 Hz to 20,000 Hz); therefore, it is called ultrasound. Appropriate oscillations at this frequency generate ultrasound used for noninvasive medical diagnoses, such as observations of a fetus in the womb.
@@ -538,14 +547,14 @@
Simple Harmonic Motion
In simple harmonic motion, the acceleration of the system, and therefore the net force, is proportional to the displacement and acts in the opposite direction of the displacement.
A good example of SHM is an object with mass m attached to a spring on a frictionless surface, as shown in Figure 15.3. The object oscillates around the equilibrium position, and the net force on the object is equal to the force provided by the spring. This force obeys Hooke’s law Fs=−kx, as discussed in a previous chapter.
+A good example of SHM is an object with mass m attached to a spring on a frictionless surface, as shown in Figure 15.3. The object oscillates around the equilibrium position, and the net force on the object is equal to the force provided by the spring. This force obeys Hooke’s law Fs=−kx, as discussed in a previous chapter.
If the net force can be described by Hooke’s law and there is no damping (slowing down due to friction or other nonconservative forces), then a simple harmonic oscillator oscillates with equal displacement on either side of the equilibrium position, as shown for an object on a spring in Figure 15.3. The maximum displacement from equilibrium is called the amplitude (A). The units for amplitude and displacement are the same but depend on the type of oscillation. For the object on the spring, the units of amplitude and displacement are meters.
+Figure 15.3 An object attached to a spring sliding on a frictionless surface is an uncomplicated simple harmonic oscillator. In the above set of figures, a mass is attached to a spring and placed on a frictionless table. The other end of the spring is attached to the wall. The position of the mass, when the spring is neither stretched nor compressed, is marked as x=0 and is the equilibrium position. (a) The mass is displaced to a position x=A and released from rest. (b) The mass accelerates as it moves in the negative x-direction, reaching a maximum negative velocity at x=0. (c) The mass continues to move in the negative x-direction, slowing until it comes to a stop at x=−A. (d) The mass now begins to accelerate in the positive x-direction, reaching a positive maximum velocity at x=0. (e) The mass then continues to move in the positive direction until it stops at x=A. The mass continues in SHM that has an amplitude A and a period T. The object’s maximum speed occurs as it passes through equilibrium. The stiffer the spring is, the smaller the period T. The greater the mass of the object is, the greater the period T.
What is so significant about SHM? For one thing, the period T and frequency f of a simple harmonic oscillator are independent of amplitude. The string of a guitar, for example, oscillates with the same frequency whether plucked gently or hard.
Two important factors do affect the period of a simple harmonic oscillator. The period is related to how stiff the system is. A very stiff object has a large force constant (k), which causes the system to have a smaller period. For example, you can adjust a diving board’s stiffness—the stiffer it is, the faster it vibrates, and the shorter its period. Period also depends on the mass of the oscillating system. The more massive the system is, the longer the period. For example, a heavy person on a diving board bounces up and down more slowly than a light one. In fact, the mass m and the force constant k are the only factors that affect the period and frequency of SHM. To derive an equation for the period and the frequency, we must first define and analyze the equations of motion. Note that the force constant is sometimes referred to as the spring constant.
@@ -553,52 +562,52 @@ Equations of SHM
-
Consider a block attached to a spring on a frictionless table (Figure 15.4). The equilibrium position (the position where the spring is neither stretched nor compressed) is marked as x=0. At the equilibrium position, the net force is zero.
+Consider a block attached to a spring on a frictionless table (Figure 15.4). The equilibrium position (the position where the spring is neither stretched nor compressed) is marked as x=0. At the equilibrium position, the net force is zero.
-Work is done on the block to pull it out to a position of x=+A, and it is then released from rest. The maximum x-position (A) is called the amplitude of the motion. The block begins to oscillate in SHM between x=+A and x=−A, where A is the amplitude of the motion and T is the period of the oscillation. The period is the time for one oscillation. Figure 15.5 shows the motion of the block as it completes one and a half oscillations after release. Figure 15.6 shows a plot of the position of the block versus time. When the position is plotted versus time, it is clear that the data can be modeled by a cosine function with an amplitude A and a period T. The cosine function cosθ repeats every multiple of 2π, whereas the motion of the block repeats every period T. However, the function cos(2πTt) repeats every integer multiple of the period. The maximum of the cosine function is one, so it is necessary to multiply the cosine function by the amplitude A.
+Figure 15.4 A block is attached to a spring and placed on a frictionless table. The equilibrium position, where the spring is neither extended nor compressed, is marked as x=0.
+Work is done on the block to pull it out to a position of x=+A, and it is then released from rest. The maximum x-position (A) is called the amplitude of the motion. The block begins to oscillate in SHM between x=+A and x=−A, where A is the amplitude of the motion and T is the period of the oscillation. The period is the time for one oscillation. Figure 15.5 shows the motion of the block as it completes one and a half oscillations after release. Figure 15.6 shows a plot of the position of the block versus time. When the position is plotted versus time, it is clear that the data can be modeled by a cosine function with an amplitude A and a period T. The cosine function cosθ repeats every multiple of 2π, whereas the motion of the block repeats every period T. However, the function cos(2πTt) repeats every integer multiple of the period. The maximum of the cosine function is one, so it is necessary to multiply the cosine function by the amplitude A.
-
x(t)=Acos(2πTt)=Acos(ωt).
(15.2)
+
x(t)=Acos(2πTt)=Acos(ωt).
(15.2)
Recall from the chapter on rotation that the angular frequency equals ω=dθdt. In this case, the period is constant, so the angular frequency is defined as 2π divided by the period, ω=2πT.
+Recall from the chapter on rotation that the angular frequency equals ω=dθdt. In this case, the period is constant, so the angular frequency is defined as 2π divided by the period, ω=2πT.
+Figure 15.5 A block is attached to one end of a spring and placed on a frictionless table. The other end of the spring is anchored to the wall. The equilibrium position, where the net force equals zero, is marked as x=0m. Work is done on the block, pulling it out to x=+A, and the block is released from rest. The block oscillates between x=+A and x=−A. The force is also shown as a vector.
-The equation for the position as a function of time x(t)=Acos(ωt) is good for modeling data, where the position of the block at the initial time t=0.00s is at the amplitude A and the initial velocity is zero. Often when taking experimental data, the position of the mass at the initial time t=0.00s is not equal to the amplitude and the initial velocity is not zero. Consider 10 seconds of data collected by a student in lab, shown in Figure 15.7.
+The equation for the position as a function of time x(t)=Acos(ωt) is good for modeling data, where the position of the block at the initial time t=0.00s is at the amplitude A and the initial velocity is zero. Often when taking experimental data, the position of the mass at the initial time t=0.00s is not equal to the amplitude and the initial velocity is not zero. Consider 10 seconds of data collected by a student in lab, shown in Figure 15.7.
-The data in Figure 15.7 can still be modeled with a periodic function, like a cosine function, but the function is shifted to the right. This shift is known as a phase shift and is usually represented by the Greek letter phi (ϕ). The equation of the position as a function of time for a block on a spring becomes
-
-This is the generalized equation for SHM where t is the time measured in seconds, ω is the angular frequency with units of inverse seconds, A is the amplitude measured in meters or centimeters, and ϕ is the phase shift measured in radians (Figure 15.8). It should be noted that because sine and cosine functions differ only by a phase shift, this motion could be modeled using either the cosine or sine function.
+Figure 15.7 Data collected by a student in lab indicate the position of a block attached to a spring, measured with a sonic range finder. The data are collected starting at time t=0.00s, but the initial position is near position x≈−0.80cm≠3.00cm, so the initial position does not equal the amplitude x0=+A. The velocity is the time derivative of the position, which is the slope at a point on the graph of position versus time. The velocity is not v=0.00m/s at time t=0.00s, as evident by the slope of the graph of position versus time, which is not zero at the initial time.
+The data in Figure 15.7 can still be modeled with a periodic function, like a cosine function, but the function is shifted to the right. This shift is known as a phase shift and is usually represented by the Greek letter phi (ϕ). The equation of the position as a function of time for a block on a spring becomes
+
+This is the generalized equation for SHM where t is the time measured in seconds, ω is the angular frequency with units of inverse seconds, A is the amplitude measured in meters or centimeters, and ϕ is the phase shift measured in radians (Figure 15.8). It should be noted that because sine and cosine functions differ only by a phase shift, this motion could be modeled using either the cosine or sine function.
+Figure 15.8 (a) A cosine function. (b) A cosine function shifted to the right by an angle ϕ. The angle ϕ is known as the phase shift of the function.
The velocity of the mass on a spring, oscillating in SHM, can be found by taking the derivative of the position equation:
-v(t)=dxdt=ddt(Acos(ωt+ϕ))=−Aωsin(ωt+φ)=−vmaxsin(ωt+ϕ).
Because the sine function oscillates between –1 and +1, the maximum velocity is the amplitude times the angular frequency, vmax=Aω. The maximum velocity occurs at the equilibrium position (x=0) when the mass is moving toward x=+A. The maximum velocity in the negative direction is attained at the equilibrium position (x=0) when the mass is moving toward x=−A and is equal to −vmax.
+v(t)=dxdt=ddt(Acos(ωt+ϕ))=−Aωsin(ωt+φ)=−vmaxsin(ωt+ϕ).
Because the sine function oscillates between –1 and +1, the maximum velocity is the amplitude times the angular frequency, vmax=Aω. The maximum velocity occurs at the equilibrium position (x=0) when the mass is moving toward x=+A. The maximum velocity in the negative direction is attained at the equilibrium position (x=0) when the mass is moving toward x=−A and is equal to −vmax.
The acceleration of the mass on the spring can be found by taking the time derivative of the velocity:
-a(t)=dvdt=ddt(−Aωsin(ωt+ϕ))=−Aω2cos(ωt+φ)=−amaxcos(ωt+ϕ).
The maximum acceleration is amax=Aω2. The maximum acceleration occurs at the position(x=−A), and the acceleration at the position (x=−A) and is equal to −amax.
+a(t)=dvdt=ddt(−Aωsin(ωt+ϕ))=−Aω2cos(ωt+φ)=−amaxcos(ωt+ϕ).
The maximum acceleration is amax=Aω2. The maximum acceleration occurs at the position(x=−A), and the acceleration at the position (x=−A) and is equal to −amax.
@@ -606,27 +615,27 @@ Summary of Equations of Motion for SHM
In summary, the oscillatory motion of a block on a spring can be modeled with the following equations of motion:
-
-
v(t)=−vmaxsin(ωt+ϕ)
(15.4)
-
a(t)=−amaxcos(ωt+ϕ)
(15.5)
-
-
-
-
Here, A is the amplitude of the motion, T is the period, ϕ is the phase shift, and ω=2πT=2πf is the angular frequency of the motion of the block.
+
+v(t)=−vmaxsin(ωt+ϕ)
(15.4)
a(t)=−amaxcos(ωt+ϕ)
(15.5)
Here, A is the amplitude of the motion, T is the period, ϕ is the phase shift, and ω=2πT=2πf is the angular frequency of the motion of the block.
Example 15.2
Determining the Equations of Motion for a Block and a Spring
-A 2.00-kg block is placed on a frictionless surface. A spring with a force constant of k=32.00N/m is attached to the block, and the opposite end of the spring is attached to the wall. The spring can be compressed or extended. The equilibrium position is marked as x=0.00m.
-
Work is done on the block, pulling it out to x=+0.02m. The block is released from rest and oscillates between x=+0.02m and x=−0.02m. The period of the motion is 1.57 s. Determine the equations of motion.
+A 2.00-kg block is placed on a frictionless surface. A spring with a force constant of
k=32.00N/m is attached to the block, and the opposite end of the spring is attached to the wall. The spring can be compressed or extended. The equilibrium position is marked as
x=0.00m.
+
Work is done on the block, pulling it out to x=+0.02m. The block is released from rest and oscillates between x=+0.02m and x=−0.02m. The period of the motion is 1.57 s. Determine the equations of motion.
Strategy
-We first find the angular frequency. The phase shift is zero, ϕ=0.00rad, because the block is released from rest at x=A=+0.02m. Once the angular frequency is found, we can determine the maximum velocity and maximum acceleration.
+We first find the angular frequency. The phase shift is zero,
ϕ=0.00rad, because the block is released from rest at
x=A=+0.02m. Once the angular frequency is found, we can determine the maximum velocity and maximum acceleration.
Solution
The angular frequency can be found and used to find the maximum velocity and maximum acceleration:
-
ωvmaxamax===2π1.57s=4.00s−1;Aω=0.02m(4.00s−1)=0.08m/s;Aω2=0.02m(4.00s−1)2=0.32m/s2.
All that is left is to fill in the equations of motion:
-
x(t)v(t)a(t)===Acos(ωt+ϕ)=(0.02m)cos(4.00s−1t);−vmaxsin(ωt+ϕ)=(−0.08m/s)sin(4.00s−1t);−amaxcos(ωt+ϕ)=(−0.32m/s2)cos(4.00s−1t).
to find equations for the angular frequency, frequency, and period.
+
One interesting characteristic of the SHM of an object attached to a spring is that the angular frequency, and therefore the period and frequency of the motion, depend on only the mass and the force constant, and not on other factors such as the amplitude of the motion. We can use the equations of motion and Newton’s second law (F⃗ net=ma⃗ ) to find equations for the angular frequency, frequency, and period.
Consider the block on a spring on a frictionless surface. There are three forces on the mass: the weight, the normal force, and the force due to the spring. The only two forces that act perpendicular to the surface are the weight and the normal force, which have equal magnitudes and opposite directions, and thus sum to zero. The only force that acts parallel to the surface is the force due to the spring, so the net force must be equal to the force of the spring:
-
Fxmamd2xdt2d2xdt2====−kx;−kx;−kx;−kmx.
Substituting the equations of motion for x and a gives us
-
−Aω2cos(ωt+ϕ)=−kmAcos(ωt+ϕ).
+
−Aω2cos(ωt+ϕ)=−kmAcos(ωt+ϕ).
Cancelling out like terms and solving for the angular frequency yields
-
The angular frequency depends only on the force constant and the mass, and not the amplitude. The angular frequency is defined as ω=2π/T, which yields an equation for the period of the motion:
+
The angular frequency depends only on the force constant and the mass, and not the amplitude. The angular frequency is defined as ω=2π/T, which yields an equation for the period of the motion:
The period also depends only on the mass and the force constant. The greater the mass, the longer the period. The stiffer the spring, the shorter the period. The frequency is
@@ -673,17 +682,17 @@
-
Figure 15.9 A spring is hung from the ceiling. When a block is attached, the block is at the equilibrium position where the weight of the block is equal to the force of the spring. (a) The spring is hung from the ceiling and the equilibrium position is marked as yo. (b) A mass is attached to the spring and a new equilibrium position is reached (y1=yo−Δy) when the force provided by the spring equals the weight of the mass. (c) The free-body diagram of the mass shows the two forces acting on the mass: the weight and the force of the spring.
-
When the block reaches the equilibrium position, as seen in Figure 15.9, the force of the spring equals the weight of the block, Fnet=Fs−mg=0, where
-
-
From the figure, the change in the position is Δy=y0−y1 and since −k(−Δy)=mg, we have
If the block is displaced and released, it will oscillate around the new equilibrium position. As shown in Figure 15.10, if the position of the block is recorded as a function of time, the recording is a periodic function.
-
If the block is displaced to a position y, the net force becomes Fnet=k(y−y0)−mg=0. But we found that at the equilibrium position, mg=kΔy=ky0−ky1. Substituting for the weight in the equation yields
-
Fnet=ky−ky0−(ky0−ky1)=−k(y−y1).
-
Recall that y1 is just the equilibrium position and any position can be set to be the point y=0.00m. So let’s set y1 to y=0.00m. The net force then becomes
-
Fnetmd2ydt2==−ky;−ky.
. (b) A mass is attached to the spring and a new equilibrium position is reached (
y1=yo−Δy) when the force provided by the spring equals the weight of the mass. (c) The free-body diagram of the mass shows the two forces acting on the mass: the weight and the force of the spring.
When the block reaches the equilibrium position, as seen in Figure 15.9, the force of the spring equals the weight of the block, Fnet=Fs−mg=0, where
+
+From the figure, the change in the position is Δy=y0−y1 and since −k(−Δy)=mg, we have
If the block is displaced and released, it will oscillate around the new equilibrium position. As shown in Figure 15.10, if the position of the block is recorded as a function of time, the recording is a periodic function.
+If the block is displaced to a position y, the net force becomes Fnet=k(y−y0)−mg=0. But we found that at the equilibrium position, mg=kΔy=ky0−ky1. Substituting for the weight in the equation yields
+Fnet=ky−ky0−(ky0−ky1)=−k(y−y1).
Recall that y1 is just the equilibrium position and any position can be set to be the point y=0.00m. So let’s set y1 to y=0.00m. The net force then becomes
+
-This is just what we found previously for a horizontally sliding mass on a spring. The constant force of gravity only served to shift the equilibrium location of the mass. Therefore, the solution should be the same form as for a block on a horizontal spring, y(t)=Acos(ωt+ϕ). The equations for the velocity and the acceleration also have the same form as for the horizontal case. Note that the inclusion of the phase shift means that the motion can actually be modeled using either a cosine or a sine function, since these two functions only differ by a phase shift.
+This is just what we found previously for a horizontally sliding mass on a spring. The constant force of gravity only served to shift the equilibrium location of the mass. Therefore, the solution should be the same form as for a block on a horizontal spring, y(t)=Acos(ωt+ϕ). The equations for the velocity and the acceleration also have the same form as for the horizontal case. Note that the inclusion of the phase shift means that the motion can actually be modeled using either a cosine or a sine function, since these two functions only differ by a phase shift.
+W=∫xixfFxdx=∫xixf−kxdx=[−12kx2]xfxi=−[12kx2f−12kx2i]=−[Uf−Ui]=−ΔU.
When considering the energy stored in a spring, the equilibrium position, marked as xi=0.00m, is the position at which the energy stored in the spring is equal to zero. When the spring is stretched or compressed a distance x, the potential energy stored in the spring is
+When considering the energy stored in a spring, the equilibrium position, marked as xi=0.00m, is the position at which the energy stored in the spring is equal to zero. When the spring is stretched or compressed a distance x, the potential energy stored in the spring is
-
+
Energy and the Simple Harmonic Oscillator
To study the energy of a simple harmonic oscillator, we need to consider all the forms of energy. Consider the example of a block attached to a spring, placed on a frictionless surface, oscillating in SHM. The potential energy stored in the deformation of the spring is
-
-In a simple harmonic oscillator, the energy oscillates between kinetic energy of the mass K=12mv2 and potential energy U=12kx2 stored in the spring. In the SHM of the mass and spring system, there are no dissipative forces, so the total energy is the sum of the potential energy and kinetic energy. In this section, we consider the conservation of energy of the system. The concepts examined are valid for all simple harmonic oscillators, including those where the gravitational force plays a role.
+
+In a simple harmonic oscillator, the energy oscillates between kinetic energy of the mass K=12mv2 and potential energy U=12kx2 stored in the spring. In the SHM of the mass and spring system, there are no dissipative forces, so the total energy is the sum of the potential energy and kinetic energy. In this section, we consider the conservation of energy of the system. The concepts examined are valid for all simple harmonic oscillators, including those where the gravitational force plays a role.
Consider Figure 15.11, which shows an oscillating block attached to a spring. In the case of undamped SHM, the energy oscillates back and forth between kinetic and potential, going completely from one form of energy to the other as the system oscillates. So for the simple example of an object on a frictionless surface attached to a spring, the motion starts with all of the energy stored in the spring as elastic potential energy. As the object starts to move, the elastic potential energy is converted into kinetic energy, becoming entirely kinetic energy at the equilibrium position. The energy is then converted back into elastic potential energy by the spring as it is stretched or compressed. The velocity becomes zero when the kinetic energy is completely converted, and this cycle then repeats. Understanding the conservation of energy in these cycles will provide extra insight here and in later applications of SHM, such as alternating circuits.
+Figure 15.11 The transformation of energy in SHM for an object attached to a spring on a frictionless surface. (a) When the mass is at the position x=+A, all the energy is stored as potential energy in the spring U=12kA2. The kinetic energy is equal to zero because the velocity of the mass is zero. (b) As the mass moves toward x=−A, the mass crosses the position x=0. At this point, the spring is neither extended nor compressed, so the potential energy stored in the spring is zero. At x=0, the total energy is all kinetic energy where K=12m(−vmax)2. (c) The mass continues to move until it reaches x=−A where the mass stops and starts moving toward x=+A. At the position x=−A, the total energy is stored as potential energy in the compressed U=12k(−A)2 and the kinetic energy is zero. (d) As the mass passes through the position x=0, the kinetic energy is K=12mv2max and the potential energy stored in the spring is zero. (e) The mass returns to the position x=+A, where K=0 and U=12kA2.
Consider Figure 15.11, which shows the energy at specific points on the periodic motion. While staying constant, the energy oscillates between the kinetic energy of the block and the potential energy stored in the spring:
-
-The motion of the block on a spring in SHM is defined by the position x(t)=Acos(ωt+ϕ) with a velocity of v(t)=−Aωsin(ωt+ϕ). Using these equations, the trigonometric identity cos2θ+sin2θ=1 and ω=km‾‾√, we can find the total energy of the system:
-ETotal=12kA2cos2(ωt+ϕ)+12mA2ω2sin2(ωt+ϕ)=12kA2cos2(ωt+ϕ)+12mA2(km)sin2(ωt+ϕ)=12kA2cos2(ωt+ϕ)+12kA2sin2(ωt+ϕ)=12kA2(cos2(ωt+ϕ)+sin2(ωt+ϕ))=12kA2.
The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude ETotal=(1/2)kA2. The total energy of the system is constant.
-A closer look at the energy of the system shows that the kinetic energy oscillates like a sine-squared function, while the potential energy oscillates like a cosine-squared function. However, the total energy for the system is constant and is proportional to the amplitude squared. Figure 15.12 shows a plot of the potential, kinetic, and total energies of the block and spring system as a function of time. Also plotted are the position and velocity as a function of time. Before time t=0.0s, the block is attached to the spring and placed at the equilibrium position. Work is done on the block by applying an external force, pulling it out to a position of x=+A. The system now has potential energy stored in the spring. At time t=0.00s, the position of the block is equal to the amplitude, the potential energy stored in the spring is equal to U=12kA2, and the force on the block is maximum and points in the negative x-direction (FS=−kA). The velocity and kinetic energy of the block are zero at time t=0.00s. At time t=0.00s, the block is released from rest.
+
+The motion of the block on a spring in SHM is defined by the position x(t)=Acos(ωt+ϕ) with a velocity of v(t)=−Aωsin(ωt+ϕ). Using these equations, the trigonometric identity cos2θ+sin2θ=1 and ω=km−−√, we can find the total energy of the system:
+ETotal=12kA2cos2(ωt+ϕ)+12mA2ω2sin2(ωt+ϕ)=12kA2cos2(ωt+ϕ)+12mA2(km)sin2(ωt+ϕ)=12kA2cos2(ωt+ϕ)+12kA2sin2(ωt+ϕ)=12kA2(cos2(ωt+ϕ)+sin2(ωt+ϕ))=12kA2.
The total energy of the system of a block and a spring is equal to the sum of the potential energy stored in the spring plus the kinetic energy of the block and is proportional to the square of the amplitude ETotal=(1/2)kA2. The total energy of the system is constant.
+A closer look at the energy of the system shows that the kinetic energy oscillates like a sine-squared function, while the potential energy oscillates like a cosine-squared function. However, the total energy for the system is constant and is proportional to the amplitude squared. Figure 15.12 shows a plot of the potential, kinetic, and total energies of the block and spring system as a function of time. Also plotted are the position and velocity as a function of time. Before time t=0.0s, the block is attached to the spring and placed at the equilibrium position. Work is done on the block by applying an external force, pulling it out to a position of x=+A. The system now has potential energy stored in the spring. At time t=0.00s, the position of the block is equal to the amplitude, the potential energy stored in the spring is equal to U=12kA2, and the force on the block is maximum and points in the negative x-direction (FS=−kA). The velocity and kinetic energy of the block are zero at time t=0.00s. At time t=0.00s, the block is released from rest.
-The potential energy curve in Figure 15.13 resembles a bowl. When a marble is placed in a bowl, it settles to the equilibrium position at the lowest point of the bowl (x=0). This happens because a restoring force points toward the equilibrium point. This equilibrium point is sometimes referred to as a fixed point. When the marble is disturbed to a different position (x=+A), the marble oscillates around the equilibrium position. Looking back at the graph of potential energy, the force can be found by looking at the slope of the potential energy graph (F=−dUdx). Since the force on either side of the fixed point points back toward the equilibrium point, the equilibrium point is called a stable equilibrium point. The points x=A and x=−A are called the turning points. (See Potential Energy and Conservation of Energy.)
+Figure 15.13 A graph of the kinetic energy (red), potential energy (blue), and total energy (green) of a simple harmonic oscillator. The force is equal to F=−dUdx. The equilibrium position is shown as a black dot and is the point where the force is equal to zero. The force is positive when x<0, negative when x>0, and equal to zero when x=0.
+The potential energy curve in Figure 15.13 resembles a bowl. When a marble is placed in a bowl, it settles to the equilibrium position at the lowest point of the bowl (x=0). This happens because a restoring force points toward the equilibrium point. This equilibrium point is sometimes referred to as a fixed point. When the marble is disturbed to a different position (x=+A), the marble oscillates around the equilibrium position. Looking back at the graph of potential energy, the force can be found by looking at the slope of the potential energy graph (F=−dUdx). Since the force on either side of the fixed point points back toward the equilibrium point, the equilibrium point is called a stable equilibrium point. The points x=A and x=−A are called the turning points. (See Potential Energy and Conservation of Energy.)
Stability is an important concept. If an equilibrium point is stable, a slight disturbance of an object that is initially at the stable equilibrium point will cause the object to oscillate around that point. The stable equilibrium point occurs because the force on either side is directed toward it. For an unstable equilibrium point, if the object is disturbed slightly, it does not return to the equilibrium point.
Consider the marble in the bowl example. If the bowl is right-side up, the marble, if disturbed slightly, will oscillate around the stable equilibrium point. If the bowl is turned upside down, the marble can be balanced on the top, at the equilibrium point where the net force is zero. However, if the marble is disturbed slightly, it will not return to the equilibrium point, but will instead roll off the bowl. The reason is that the force on either side of the equilibrium point is directed away from that point. This point is an unstable equilibrium point.
Figure 15.14 shows three conditions. The first is a stable equilibrium point (a), the second is an unstable equilibrium point (b), and the last is also an unstable equilibrium point (c), because the force on only one side points toward the equilibrium point.
@@ -823,7 +832,7 @@ The process of determining whether an equilibrium point is stable or unstable can be formalized. Consider the potential energy curves shown in Figure 15.15. The force can be found by analyzing the slope of the graph. The force is F=−dUdx. In (a), the fixed point is at x=0.00m. When x<0.00m, the force is positive. When x>0.00m, the force is negative. This is a stable point. In (b), the fixed point is at x=0.00m. When x<0.00m, the force is negative. When x>0.00m, the force is also negative. This is an unstable point.
+The process of determining whether an equilibrium point is stable or unstable can be formalized. Consider the potential energy curves shown in Figure 15.15. The force can be found by analyzing the slope of the graph. The force is F=−dUdx. In (a), the fixed point is at x=0.00m. When x<0.00m, the force is positive. When x>0.00m, the force is negative. This is a stable point. In (b), the fixed point is at x=0.00m. When x<0.00m, the force is negative. When x>0.00m, the force is also negative. This is an unstable point.
A practical application of the concept of stable equilibrium points is the force between two neutral atoms in a molecule. If two molecules are in close proximity, separated by a few atomic diameters, they can experience an attractive force. If the molecules move close enough so that the electron shells of the other electrons overlap, the force between the molecules becomes repulsive. The attractive force between the two atoms may cause the atoms to form a molecule. The force between the two molecules is not a linear force and cannot be modeled simply as two masses separated by a spring, but the atoms of the molecule can oscillate around an equilibrium point when displaced a small amount from the equilibrium position. The atoms oscillate due the attractive force and repulsive force between the two atoms.
Consider one example of the interaction between two atoms known as the van Der Waals interaction. It is beyond the scope of this chapter to discuss in depth the interactions of the two atoms, but the oscillations of the atoms can be examined by considering one example of a model of the potential energy of the system. One suggestion to model the potential energy of this molecule is with the Lennard-Jones 6-12 potential:
-
-A graph of this function is shown in Figure 15.16. The two parameters ε and σ are found experimentally.
+
+A graph of this function is shown in Figure 15.16. The two parameters ε and σ are found experimentally.
-From the graph, you can see that there is a potential energy well, which has some similarities to the potential energy well of the potential energy function of the simple harmonic oscillator discussed in Figure 15.13. The Lennard-Jones potential has a stable equilibrium point where the potential energy is minimum and the force on either side of the equilibrium point points toward equilibrium point. Note that unlike the simple harmonic oscillator, the potential well of the Lennard-Jones potential is not symmetric. This is due to the fact that the force between the atoms is not a Hooke’s law force and is not linear. The atoms can still oscillate around the equilibrium position xmin because when x<xmin, the force is positive; when x>xmin, the force is negative. Notice that as x approaches zero, the slope is quite steep and negative, which means that the force is large and positive. This suggests that it takes a large force to try to push the atoms close together. As x becomes increasingly large, the slope becomes less steep and the force is smaller and negative. This suggests that if given a large enough energy, the atoms can be separated.
-If you are interested in this interaction, find the force between the molecules by taking the derivative of the potential energy function. You will see immediately that the force does not resemble a Hooke’s law force (F=−kx), but if you are familiar with the binomial theorem:
-(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯,
From the graph, you can see that there is a potential energy well, which has some similarities to the potential energy well of the potential energy function of the simple harmonic oscillator discussed in Figure 15.13. The Lennard-Jones potential has a stable equilibrium point where the potential energy is minimum and the force on either side of the equilibrium point points toward equilibrium point. Note that unlike the simple harmonic oscillator, the potential well of the Lennard-Jones potential is not symmetric. This is due to the fact that the force between the atoms is not a Hooke’s law force and is not linear. The atoms can still oscillate around the equilibrium position xmin because when x<xmin, the force is positive; when x>xmin, the force is negative. Notice that as x approaches zero, the slope is quite steep and negative, which means that the force is large and positive. This suggests that it takes a large force to try to push the atoms close together. As x becomes increasingly large, the slope becomes less steep and the force is smaller and negative. This suggests that if given a large enough energy, the atoms can be separated.
+If you are interested in this interaction, find the force between the molecules by taking the derivative of the potential energy function. You will see immediately that the force does not resemble a Hooke’s law force (F=−kx), but if you are familiar with the binomial theorem:
+(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯,
the force can be approximated by a Hooke’s law force.
Velocity and Energy Conservation
-Getting back to the system of a block and a spring in Figure 15.11, once the block is released from rest, it begins to move in the negative direction toward the equilibrium position. The potential energy decreases and the magnitude of the velocity and the kinetic energy increase. At time t=T/4, the block reaches the equilibrium position x=0.00m, where the force on the block and the potential energy are zero. At the equilibrium position, the block reaches a negative velocity with a magnitude equal to the maximum velocity v=−Aω. The kinetic energy is maximum and equal to K=12mv2=12mA2ω2=12kA2. At this point, the force on the block is zero, but momentum carries the block, and it continues in the negative direction toward x=−A. As the block continues to move, the force on it acts in the positive direction and the magnitude of the velocity and kinetic energy decrease. The potential energy increases as the spring compresses. At time t=T/2, the block reaches x=−A. Here the velocity and kinetic energy are equal to zero. The force on the block is F=+kA and the potential energy stored in the spring is U=12kA2. During the oscillations, the total energy is constant and equal to the sum of the potential energy and the kinetic energy of the system,
+Getting back to the system of a block and a spring in Figure 15.11, once the block is released from rest, it begins to move in the negative direction toward the equilibrium position. The potential energy decreases and the magnitude of the velocity and the kinetic energy increase. At time t=T/4, the block reaches the equilibrium position x=0.00m, where the force on the block and the potential energy are zero. At the equilibrium position, the block reaches a negative velocity with a magnitude equal to the maximum velocity v=−Aω. The kinetic energy is maximum and equal to K=12mv2=12mA2ω2=12kA2. At this point, the force on the block is zero, but momentum carries the block, and it continues in the negative direction toward x=−A. As the block continues to move, the force on it acts in the positive direction and the magnitude of the velocity and kinetic energy decrease. The potential energy increases as the spring compresses. At time t=T/2, the block reaches x=−A. Here the velocity and kinetic energy are equal to zero. The force on the block is F=+kA and the potential energy stored in the spring is U=12kA2. During the oscillations, the total energy is constant and equal to the sum of the potential energy and the kinetic energy of the system,
-
ETotal=12kx2+12mv2=12kA2.
(15.12)
+
ETotal=12kx2+12mv2=12kA2.
(15.12)
The equation for the energy associated with SHM can be solved to find the magnitude of the velocity at any position:
-
|v|=km(A2−x2)‾‾‾‾‾‾‾‾‾‾‾‾√.
(15.13)
+
|v|=km(A2−x2)−−−−−−−−−−√.
(15.13)
The energy in a simple harmonic oscillator is proportional to the square of the amplitude. When considering many forms of oscillations, you will find the energy proportional to the amplitude squared.
Check Your Understanding 15.1
@@ -942,7 +951,7 @@
An easy way to model SHM is by considering uniform circular motion. Figure 15.17 shows one way of using this method. A peg (a cylinder of wood) is attached to a vertical disk, rotating with a constant angular frequency. Figure 15.18 shows a side view of the disk and peg. If a lamp is placed above the disk and peg, the peg produces a shadow. Let the disk have a radius of r=A and define the position of the shadow that coincides with the center line of the disk to be x=0.00m. As the disk rotates at a constant rate, the shadow oscillates between x=+A and x=−A. Now imagine a block on a spring beneath the floor as shown in Figure 15.18.
+
An easy way to model SHM is by considering uniform circular motion. Figure 15.17 shows one way of using this method. A peg (a cylinder of wood) is attached to a vertical disk, rotating with a constant angular frequency. Figure 15.18 shows a side view of the disk and peg. If a lamp is placed above the disk and peg, the peg produces a shadow. Let the disk have a radius of r=A and define the position of the shadow that coincides with the center line of the disk to be x=0.00m. As the disk rotates at a constant rate, the shadow oscillates between x=+A and x=−A. Now imagine a block on a spring beneath the floor as shown in Figure 15.18.
@@ -960,7 +969,7 @@
-
Figure 15.18 Light shines down on the disk so that the peg makes a shadow. If the disk rotates at just the right angular frequency, the shadow follows the motion of the block on a spring. If there is no energy dissipated due to nonconservative forces, the block and the shadow will oscillate back and forth in unison. In this figure, four snapshots are taken at four different times. (a) The wheel starts at θ=0o and the shadow of the peg is at x=+A, representing the mass at position x=+A. (b) As the disk rotates through an angle θ=ωt, the shadow of the peg is between x=+A and x=0. (c) The disk continues to rotate until θ=900, at which the shadow follows the mass to x=0. (d) The disk continues to rotate, the shadow follows the position of the mass.
+
Figure 15.18 Light shines down on the disk so that the peg makes a shadow. If the disk rotates at just the right angular frequency, the shadow follows the motion of the block on a spring. If there is no energy dissipated due to nonconservative forces, the block and the shadow will oscillate back and forth in unison. In this figure, four snapshots are taken at four different times. (a) The wheel starts at θ=0o and the shadow of the peg is at x=+A, representing the mass at position x=+A. (b) As the disk rotates through an angle θ=ωt, the shadow of the peg is between x=+A and x=0. (c) The disk continues to rotate until θ=900, at which the shadow follows the mass to x=0. (d) The disk continues to rotate, the shadow follows the position of the mass.
-
+
-
Recall that the block attached to the spring does not move at a constant velocity. How often does the wheel have to turn to have the peg’s shadow always on the block? The disk must turn at a constant angular frequency equal to 2π times the frequency of oscillation (ω=2πf).
+Recall that the block attached to the spring does not move at a constant velocity. How often does the wheel have to turn to have the peg’s shadow always on the block? The disk must turn at a constant angular frequency equal to 2π times the frequency of oscillation (ω=2πf).
-Figure 15.19 shows the basic relationship between uniform circular motion and SHM. The peg lies at the tip of the radius, a distance A from the center of the disk. The x-axis is defined by a line drawn parallel to the ground, cutting the disk in half. The y-axis (not shown) is defined by a line perpendicular to the ground, cutting the disk into a left half and a right half. The center of the disk is the point (x=0,y=0). The projection of the position of the peg onto the fixed x-axis gives the position of the shadow, which undergoes SHM analogous to the system of the block and spring. At the time shown in the figure, the projection has position x and moves to the left with velocity v. The tangential velocity of the peg around the circle equals v–max of the block on the spring. The x-component of the velocity is equal to the velocity of the block on the spring.
+Figure 15.19 shows the basic relationship between uniform circular motion and SHM. The peg lies at the tip of the radius, a distance A from the center of the disk. The x-axis is defined by a line drawn parallel to the ground, cutting the disk in half. The y-axis (not shown) is defined by a line perpendicular to the ground, cutting the disk into a left half and a right half. The center of the disk is the point (x=0,y=0). The projection of the position of the peg onto the fixed x-axis gives the position of the shadow, which undergoes SHM analogous to the system of the block and spring. At the time shown in the figure, the projection has position x and moves to the left with velocity v. The tangential velocity of the peg around the circle equals v–max of the block on the spring. The x-component of the velocity is equal to the velocity of the block on the spring.
@@ -987,16 +996,16 @@
-Figure 15.19 A peg moving on a circular path with a constant angular velocity ω is undergoing uniform circular motion. Its projection on the x-axis undergoes SHM. Also shown is the velocity of the peg around the circle, vmax, and its projection, which is v. Note that these velocities form a similar triangle to the displacement triangle.
+Figure 15.19 A peg moving on a circular path with a constant angular velocity ω is undergoing uniform circular motion. Its projection on the x-axis undergoes SHM. Also shown is the velocity of the peg around the circle, vmax, and its projection, which is v. Note that these velocities form a similar triangle to the displacement triangle.
-We can use Figure 15.19 to analyze the velocity of the shadow as the disk rotates. The peg moves in a circle with a speed of vmax=Aω. The shadow moves with a velocity equal to the component of the peg’s velocity that is parallel to the surface where the shadow is being produced:
+We can use Figure 15.19 to analyze the velocity of the shadow as the disk rotates. The peg moves in a circle with a speed of vmax=Aω. The shadow moves with a velocity equal to the component of the peg’s velocity that is parallel to the surface where the shadow is being produced:
@@ -1137,7 +1146,7 @@ Figure 15.25 shows a mass m attached to a spring with a force constant k. The mass is raised to a position A0, the initial amplitude, and then released. The mass oscillates around the equilibrium position in a fluid with viscosity but the amplitude decreases for each oscillation. For a system that has a small amount of damping, the period and frequency are constant and are nearly the same as for SHM, but the amplitude gradually decreases as shown. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy.
+
Figure 15.25 shows a mass m attached to a spring with a force constant k. The mass is raised to a position A0, the initial amplitude, and then released. The mass oscillates around the equilibrium position in a fluid with viscosity but the amplitude decreases for each oscillation. For a system that has a small amount of damping, the period and frequency are constant and are nearly the same as for SHM, but the amplitude gradually decreases as shown. This occurs because the non-conservative damping force removes energy from the system, usually in the form of thermal energy.
@@ -1150,11 +1159,11 @@ Consider the forces acting on the mass. Note that the only contribution of the weight is to change the equilibrium position, as discussed earlier in the chapter. Therefore, the net force is equal to the force of the spring and the damping force (FD). If the magnitude of the velocity is small, meaning the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion (FD=−bv). The net force on the mass is therefore
+
Consider the forces acting on the mass. Note that the only contribution of the weight is to change the equilibrium position, as discussed earlier in the chapter. Therefore, the net force is equal to the force of the spring and the damping force (FD). If the magnitude of the velocity is small, meaning the mass oscillates slowly, the damping force is proportional to the velocity and acts against the direction of motion (FD=−bv). The net force on the mass is therefore
-
+
@@ -1163,17 +1172,17 @@
-md2xdt2+bdxdt+kx=0.
(15.23)
md2xdt2+bdxdt+kx=0.
(15.23)
To determine the solution to this equation, consider the plot of position versus time shown in Figure 15.26. The curve resembles a cosine curve oscillating in the envelope of an exponential function A0e−αt where α=b2m. The solution is
+To determine the solution to this equation, consider the plot of position versus time shown in Figure 15.26. The curve resembles a cosine curve oscillating in the envelope of an exponential function A0e−αt where α=b2m. The solution is
-
x(t)=A0e−b2mtcos(ωt+ϕ).
(15.24)
+
x(t)=A0e−b2mtcos(ωt+ϕ).
(15.24)
ω=km−(b2m)2‾‾‾‾‾‾‾‾‾‾‾‾‾√.
+
@@ -1191,7 +1200,7 @@
-
+
@@ -1201,7 +1210,7 @@
-ω=ω20−(b2m)2‾‾‾‾‾‾‾‾‾‾‾‾‾√.
(15.26)
ω=ω20−(b2m)2−−−−−−−−−−−√.
(15.26)
ω=km−(b2m)2‾‾‾‾‾‾‾‾‾‾‾‾‾√.
+
-As b increases, km−(b2m)2 becomes smaller and eventually reaches zero when b=4mk‾‾‾‾√. If b becomes any larger, km−(b2m)2 becomes a negative number and km−(b2m)2‾‾‾‾‾‾‾‾‾‾‾√ is a complex number.
+As b increases, km−(b2m)2 becomes smaller and eventually reaches zero when b=4mk−−−−√. If b becomes any larger, km−(b2m)2 becomes a negative number and km−(b2m)2−−−−−−−−−−√ is a complex number.
-Figure 15.27 shows the displacement of a harmonic oscillator for different amounts of damping. When the damping constant is small, b<4mk‾‾‾‾√, the system oscillates while the amplitude of the motion decays exponentially. This system is said to be underdamped, as in curve (a). Many systems are underdamped, and oscillate while the amplitude decreases exponentially, such as the mass oscillating on a spring. The damping may be quite small, but eventually the mass comes to rest. If the damping constant is b=4mk‾‾‾‾√, the system is said to be critically damped, as in curve (b). An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible. Curve (c) in Figure 15.27 represents an overdamped system where b>4mk‾‾‾‾√. An overdamped system will approach equilibrium over a longer period of time.
+Figure 15.27 shows the displacement of a harmonic oscillator for different amounts of damping. When the damping constant is small, b<4mk−−−−√, the system oscillates while the amplitude of the motion decays exponentially. This system is said to be underdamped, as in curve (a). Many systems are underdamped, and oscillate while the amplitude decreases exponentially, such as the mass oscillating on a spring. The damping may be quite small, but eventually the mass comes to rest. If the damping constant is b=4mk−−−−√, the system is said to be critically damped, as in curve (b). An example of a critically damped system is the shock absorbers in a car. It is advantageous to have the oscillations decay as fast as possible. Here, the system does not oscillate, but asymptotically approaches the equilibrium condition as quickly as possible. Curve (c) in Figure 15.27 represents an overdamped system where b>4mk−−−−√. An overdamped system will approach equilibrium over a longer period of time.
@@ -1236,7 +1245,7 @@
-Figure 15.27 The position versus time for three systems consisting of a mass and a spring in a viscous fluid. (a) If the damping is small (b<4mk‾‾‾‾√), the mass oscillates, slowly losing amplitude as the energy is dissipated by the non-conservative force(s). The limiting case is (b) where the damping is (b=4mk‾‾‾‾√). (c) If the damping is very large (b>4mk‾‾‾‾√), the mass does not oscillate when displaced, but attempts to return to the equilibrium position.
+Figure 15.27 The position versus time for three systems consisting of a mass and a spring in a viscous fluid. (a) If the damping is small (b<4mk−−−−√), the mass oscillates, slowly losing amplitude as the energy is dissipated by the non-conservative force(s). The limiting case is (b) where the damping is (b=4mk−−−−√). (c) If the damping is very large (b>4mk−−−−√), the mass does not oscillate when displaced, but attempts to return to the equilibrium position.
@@ -1342,11 +1351,11 @@
-Figure 15.29 The paddle ball on its rubber band moves in response to the finger supporting it. If the finger moves with the natural frequency f0 of the ball on the rubber band, then a resonance is achieved, and the amplitude of the ball’s oscillations increases dramatically. At higher and lower driving frequencies, energy is transferred to the ball less efficiently, and it responds with lower-amplitude oscillations.
+Figure 15.29 The paddle ball on its rubber band moves in response to the finger supporting it. If the finger moves with the natural frequency f0 of the ball on the rubber band, then a resonance is achieved, and the amplitude of the ball’s oscillations increases dramatically. At higher and lower driving frequencies, energy is transferred to the ball less efficiently, and it responds with lower-amplitude oscillations.
-Consider a simple experiment. Attach a mass m to a spring in a viscous fluid, similar to the apparatus discussed in the damped harmonic oscillator. This time, instead of fixing the free end of the spring, attach the free end to a disk that is driven by a variable-speed motor. The motor turns with an angular driving frequency of ω. The rotating disk provides energy to the system by the work done by the driving force (Fd=F0sin(ωt)). The experimental apparatus is shown in Figure 15.30.
+Consider a simple experiment. Attach a mass m to a spring in a viscous fluid, similar to the apparatus discussed in the damped harmonic oscillator. This time, instead of fixing the free end of the spring, attach the free end to a disk that is driven by a variable-speed motor. The motor turns with an angular driving frequency of ω. The rotating disk provides energy to the system by the work done by the driving force (Fd=F0sin(ωt)). The experimental apparatus is shown in Figure 15.30.
@@ -1359,12 +1368,12 @@ Using Newton’s second law (F⃗ net=ma⃗ ), we can analyze the motion of the mass. The resulting equation is similar to the force equation for the damped harmonic oscillator, with the addition of the driving force:
+
Using Newton’s second law (F⃗ net=ma⃗ ), we can analyze the motion of the mass. The resulting equation is similar to the force equation for the damped harmonic oscillator, with the addition of the driving force:
-
−kx−bdxdt+F0sin(ωt)=md2xdt2.
(15.27)
+
−kx−bdxdt+F0sin(ωt)=md2xdt2.
(15.27)
-
+
-
Once again, it is left as an exercise to prove that this equation is a solution. Taking the first and second time derivative of x(t) and substituting them into the force equation shows that x(t)=Asin(ωt+ϕ) is a solution as long as the amplitude is equal to
+Once again, it is left as an exercise to prove that this equation is a solution. Taking the first and second time derivative of x(t) and substituting them into the force equation shows that x(t)=Asin(ωt+ϕ) is a solution as long as the amplitude is equal to
-
A=F0m(ω2−ω20)2+b2ω2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
(15.29)
+
A=F0m(ω2−ω20)2+b2ω2−−−−−−−−−−−−−−−−√
(15.29)
where ω0=km‾‾√ is the natural angular frequency of the system of the mass and spring. Recall that the angular frequency, and therefore the frequency, of the motor can be adjusted. Looking at the denominator of the equation for the amplitude, when the driving frequency is much smaller, or much larger, than the natural frequency, the square of the difference of the two angular frequencies (ω2−ω20)2 is positive and large, making the denominator large, and the result is a small amplitude for the oscillations of the mass. As the frequency of the driving force approaches the natural frequency of the system, the denominator becomes small and the amplitude of the oscillations becomes large. The maximum amplitude results when the frequency of the driving force equals the natural frequency of the system (Amax=F0bω).
+where ω0=km−−√ is the natural angular frequency of the system of the mass and spring. Recall that the angular frequency, and therefore the frequency, of the motor can be adjusted. Looking at the denominator of the equation for the amplitude, when the driving frequency is much smaller, or much larger, than the natural frequency, the square of the difference of the two angular frequencies (ω2−ω20)2 is positive and large, making the denominator large, and the result is a small amplitude for the oscillations of the mass. As the frequency of the driving force approaches the natural frequency of the system, the denominator becomes small and the amplitude of the oscillations becomes large. The maximum amplitude results when the frequency of the driving force equals the natural frequency of the system (Amax=F0bω).
-Figure 15.31 shows a graph of the amplitude of a damped harmonic oscillator as a function of the frequency of the periodic force driving it. Each of the three curves on the graph represents a different amount of damping. All three curves peak at the point where the frequency of the driving force equals the natural frequency of the harmonic oscillator. The highest peak, or greatest response, is for the least amount of damping, because less energy is removed by the damping force. Note that since the amplitude grows as the damping decreases, taking this to the limit where there is no damping (b=0), the amplitude becomes infinite.
+Figure 15.31 shows a graph of the amplitude of a damped harmonic oscillator as a function of the frequency of the periodic force driving it. Each of the three curves on the graph represents a different amount of damping. All three curves peak at the point where the frequency of the driving force equals the natural frequency of the harmonic oscillator. The highest peak, or greatest response, is for the least amount of damping, because less energy is removed by the damping force. Note that since the amplitude grows as the damping decreases, taking this to the limit where there is no damping (b=0), the amplitude becomes infinite.
@@ -1410,7 +1419,7 @@ It is interesting to note that the widths of the resonance curves shown in Figure 15.31 depend on damping: the less the damping, the narrower the resonance. The consequence is that if you want a driven oscillator to resonate at a very specific frequency, you need as little damping as possible. For instance, a radio has a circuit that is used to choose a particular radio station. In this case, the forced damped oscillator consists of a resistor, capacitor, and inductor, which will be discussed later in this course. The circuit is “tuned” to pick a particular radio station. Here it is desirable to have the resonance curve be very narrow, to pick out the exact frequency of the radio station chosen. The narrowness of the graph, and the ability to pick out a certain frequency, is known as the quality of the system. The quality is defined as the spread of the angular frequency, or equivalently, the spread in the frequency, at half the maximum amplitude, divided by the natural frequency (Q=Δωω0) as shown in Figure 15.32. For a small damping, the quality is approximately equal to Q≈2bm.
+
It is interesting to note that the widths of the resonance curves shown in Figure 15.31 depend on damping: the less the damping, the narrower the resonance. The consequence is that if you want a driven oscillator to resonate at a very specific frequency, you need as little damping as possible. For instance, a radio has a circuit that is used to choose a particular radio station. In this case, the forced damped oscillator consists of a resistor, capacitor, and inductor, which will be discussed later in this course. The circuit is “tuned” to pick a particular radio station. Here it is desirable to have the resonance curve be very narrow, to pick out the exact frequency of the radio station chosen. The narrowness of the graph, and the ability to pick out a certain frequency, is known as the quality of the system. The quality is defined as the spread of the angular frequency, or equivalently, the spread in the frequency, at half the maximum amplitude, divided by the natural frequency (Q=Δωω0) as shown in Figure 15.32. For a small damping, the quality is approximately equal to Q≈2bm.
@@ -1586,149 +1595,149 @@ Chapter Review
| Relationship between frequency and period |
-f=1T |
+f=1T |
-| Position in SHM withϕ=0.00 |
-x(t)=Acos(ωt) |
+Position in SHM withϕ=0.00 |
+x(t)=Acos(ωt) |
| General position in SHM |
-x(t)=Acos(ωt+ϕ) |
+x(t)=Acos(ωt+ϕ) |
| General velocity in SHM |
-v(t)=−Aωsin(ωt+ϕ) |
+v(t)=−Aωsin(ωt+ϕ) |
| General acceleration in SHM |
-a(t)=−Aω2cos(ωt+ϕ) |
+a(t)=−Aω2cos(ωt+ϕ) |
| Maximum displacement (amplitude) of SHM |
-xmax=A |
+xmax=A |
| Maximum velocity of SHM |
-|vmax|=Aω |
+|vmax|=Aω |
| Maximum acceleration of SHM |
-|amax|=Aω2 |
+|amax|=Aω2 |
| Angular frequency of a mass-spring system in SHM |
-ω=km‾‾√ |
+ω=km−−√ |
| Period of a mass-spring system in SHM |
-T=2πmk‾‾√ |
+T=2πmk−−√ |
| Frequency of a mass-spring system in SHM |
-f=12πkm‾‾√ |
+f=12πkm−−√ |
| Energy in a mass-spring system in SHM |
-ETotal=12kx2+12mv2=12kA2 |
+ETotal=12kx2+12mv2=12kA2 |
| The velocity of the mass in a spring-mass
system in SHM |
-v=±km(A2−x2)‾‾‾‾‾‾‾‾‾‾‾√ |
+v=±km(A2−x2)−−−−−−−−−√ |
| The x-component of the radius of a rotating disk |
-x(t)=Acos(ωt+ϕ) |
+x(t)=Acos(ωt+ϕ) |
| The x-component of the velocity of the edge of a rotating disk |
-v(t)=−vmaxsin(ωt+ϕ) |
+v(t)=−vmaxsin(ωt+ϕ) |
| The x-component of the acceleration of the
edge of a rotating disk |
-a(t)=−amaxcos(ωt+ϕ) |
+a(t)=−amaxcos(ωt+ϕ) |
| Force equation for a simple pendulum |
-d2θdt2=−gLθ |
+d2θdt2=−gLθ |
| Angular frequency for a simple pendulum |
-ω=gL‾‾√ |
+ω=gL−−√ |
| Period of a simple pendulum |
-T=2πLg‾‾√ |
+T=2πLg−−√ |
| Angular frequency of a physical pendulum |
-ω=mgLI‾‾‾‾√ |
+ω=mgLI−−−−√ |
| Period of a physical pendulum |
-T=2πImgL‾‾‾‾√ |
+T=2πImgL−−−−√ |
| Period of a torsional pendulum |
-T=2πIκ‾‾√ |
+T=2πIκ−−√ |
| Newton’s second law for harmonic motion |
-md2xdt2+bdxdt+kx=0 |
+md2xdt2+bdxdt+kx=0 |
| Solution for underdamped harmonic motion |
-x(t)=A0e−b2mtcos(ωt+ϕ) |
+x(t)=A0e−b2mtcos(ωt+ϕ) |
| Natural angular frequency of a
mass-spring system |
-ω0=km‾‾√ |
+ω0=km−−√ |
| Angular frequency of underdamped
harmonic motion |
-ω=ω20−(b2m)2‾‾‾‾‾‾‾‾‾‾‾√ |
+ω=ω20−(b2m)2−−−−−−−−−−√ |
| Newton’s second law for forced,
damped oscillation |
-−kx−bdxdt+Fosin(ωt)=md2xdt2 |
+−kx−bdxdt+Fosin(ωt)=md2xdt2 |
| Solution to Newton’s second law for forced,
damped oscillations |
-x(t)=Acos(ωt+ϕ) |
+x(t)=Acos(ωt+ϕ) |
| Amplitude of system undergoing forced,
damped oscillations |
-A=Fom(ω2−ω2o)2+b2ω2√ |
+A=Fom(ω2−ω2o)2+b2ω2√ |
Summary
15.1 Simple Harmonic Motion
-- Periodic motion is a repeating oscillation. The time for one oscillation is the period T and the number of oscillations per unit time is the frequency f. These quantities are related by f=1T.
+- Periodic motion is a repeating oscillation. The time for one oscillation is the period T and the number of oscillations per unit time is the frequency f. These quantities are related by f=1T.
- Simple harmonic motion (SHM) is oscillatory motion for a system where the restoring force is proportional to the displacement and acts in the direction opposite to the displacement.
-- Maximum displacement is the amplitude A. The angular frequency ω, period T, and frequency f of a simple harmonic oscillator are given by ω=km‾‾√, T=2πmk‾‾√,andf=12πkm‾‾√, where m is the mass of the system and k is the force constant.
-- Displacement as a function of time in SHM is given byx(t)=Acos(2πTt+ϕ)=Acos(ωt+ϕ).
-- The velocity is given by v(t)=−Aωsin(ωt+ϕ)=−vmaxsin(ωt+ϕ),wherevmax=Aω=Akm‾‾√.
-- The acceleration is a(t)=−Aω2cos(ωt+ϕ)=−amaxcos(ωt+ϕ), where amax=Aω2=Akm.
+- Maximum displacement is the amplitude A. The angular frequency ω, period T, and frequency f of a simple harmonic oscillator are given by ω=km−−√, T=2πmk−−√,andf=12πkm−−√, where m is the mass of the system and k is the force constant.
+- Displacement as a function of time in SHM is given byx(t)=Acos(2πTt+ϕ)=Acos(ωt+ϕ).
+- The velocity is given by v(t)=−Aωsin(ωt+ϕ)=−vmaxsin(ωt+ϕ),wherevmax=Aω=Akm−−√.
+- The acceleration is a(t)=−Aω2cos(ωt+ϕ)=−amaxcos(ωt+ϕ), where amax=Aω2=Akm.
15.2 Energy in Simple Harmonic Motion
15.3 Comparing Simple Harmonic Motion and Circular Motion
- A projection of uniform circular motion undergoes simple harmonic oscillation.
-- Consider a circle with a radius A, moving at a constant angular speed ω. A point on the edge of the circle moves at a constant tangential speed of vmax=Aω. The projection of the radius onto the x-axis is x(t)=Acos(ωt+ϕ), where (ϕ) is the phase shift. The x-component of the tangential velocity is v(t)=−Aωsin(ωt+ϕ).
+- Consider a circle with a radius A, moving at a constant angular speed ω. A point on the edge of the circle moves at a constant tangential speed of vmax=Aω. The projection of the radius onto the x-axis is x(t)=Acos(ωt+ϕ), where (ϕ) is the phase shift. The x-component of the tangential velocity is v(t)=−Aωsin(ωt+ϕ).
15.4 Pendulums
-- A mass m suspended by a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than about 15°. The period of a simple pendulum is T=2πLg‾‾√, where L is the length of the string and g is the acceleration due to gravity.
-- The period of a physical pendulum T=2πImgL‾‾‾‾√ can be found if the moment of inertia is known. The length between the point of rotation and the center of mass is L.
-- The period of a torsional pendulum T=2πIκ‾‾√ can be found if the moment of inertia and torsion constant are known.
+- A mass m suspended by a wire of length L and negligible mass is a simple pendulum and undergoes SHM for amplitudes less than about 15°. The period of a simple pendulum is T=2πLg−−√, where L is the length of the string and g is the acceleration due to gravity.
+- The period of a physical pendulum T=2πImgL−−−−√ can be found if the moment of inertia is known. The length between the point of rotation and the center of mass is L.
+- The period of a torsional pendulum T=2πIκ−−√ can be found if the moment of inertia and torsion constant are known.
15.5 Damped Oscillations
@@ -1865,7 +1874,7 @@ Chapter Review
Problems
15.1 Simple Harmonic Motion
21.
-
Prove that using x(t)=Asin(ωt+ϕ) will produce the same results for the period for the oscillations of a mass and a spring. Why do you think the cosine function was chosen?
+
Prove that using x(t)=Asin(ωt+ϕ) will produce the same results for the period for the oscillations of a mass and a spring. Why do you think the cosine function was chosen?
@@ -1882,12 +1891,12 @@ Chapter Review
24.
-
Find the frequency of a tuning fork that takes 2.50×10−3s to complete one oscillation.
+
Find the frequency of a tuning fork that takes 2.50×10−3s to complete one oscillation.
25.
-
A stroboscope is set to flash every 8.00×10−5s. What is the frequency of the flashes?
+
A stroboscope is set to flash every 8.00×10−5s. What is the frequency of the flashes?
@@ -1909,7 +1918,7 @@
Chapter Review
29.
-
A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and released. The mass oscillates with a frequency f0. If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of f0
+
A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and released. The mass oscillates with a frequency f0. If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of f0
?
@@ -1955,14 +1964,14 @@
Chapter Review
37.
-
The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40×104N/m. (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? (Hint: Use conservation of energy.) (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.
+
The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40×104N/m. (a) What is the frequency at which he bounces, given his mass plus and the mass of his equipment are 90.0 kg? (b) How much would this rope stretch to break the climber’s fall if he free-falls 2.00 m before the rope runs out of slack? (Hint: Use conservation of energy.) (c) Repeat both parts of this problem in the situation where twice this length of nylon rope is used.
15.3 Comparing Simple Harmonic Motion and Circular Motion
38.
-
The motion of a mass on a spring hung vertically, where the mass oscillates up and down, can also be modeled using the rotating disk. Instead of the lights being placed horizontally along the top and pointing down, place the lights vertically and have the lights shine on the side of the rotating disk. A shadow will be produced on a nearby wall, and will move up and down. Write the equations of motion for the shadow taking the position at t=0.0s to be y=0.0m with the mass moving in the positive y-direction.
+
The motion of a mass on a spring hung vertically, where the mass oscillates up and down, can also be modeled using the rotating disk. Instead of the lights being placed horizontally along the top and pointing down, place the lights vertically and have the lights shine on the side of the rotating disk. A shadow will be produced on a nearby wall, and will move up and down. Write the equations of motion for the shadow taking the position at t=0.0s to be y=0.0m with the mass moving in the positive y-direction.
@@ -2022,12 +2031,12 @@
Chapter Review
48.
-
(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79m/s2 is moved to a location where the acceleration due to gravity is 9.82m/s2. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
+
(a) A pendulum that has a period of 3.00000 s and that is located where the acceleration due to gravity is 9.79m/s2 is moved to a location where the acceleration due to gravity is 9.82m/s2. What is its new period? (b) Explain why so many digits are needed in the value for the period, based on the relation between the period and the acceleration due to gravity.
49.
-
A pendulum with a period of 2.00000 s in one location (g=9.80m/s2) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
+
A pendulum with a period of 2.00000 s in one location (g=9.80m/s2) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?
@@ -2039,7 +2048,7 @@ Chapter Review
15.5 Damped Oscillations
51.
-
The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
+
The amplitude of a lightly damped oscillator decreases by 3.0% during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?
@@ -2051,7 +2060,7 @@ Chapter Review
53.
-
If a car has a suspension system with a force constant of 5.00×104N/m, how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?
+
If a car has a suspension system with a force constant of 5.00×104N/m, how much energy must the car’s shocks remove to dampen an oscillation starting with a maximum displacement of 0.0750 m?
@@ -2062,14 +2071,14 @@ Chapter Review
55.
-
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction μs=0.100. (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is μk=0.0850, what total distance does it travel before stopping? Assume it starts at the maximum amplitude.
+
Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is simple friction between the object and surface with a static coefficient of friction μs=0.100. (a) How far can the spring be stretched without moving the mass? (b) If the object is set into oscillation with an amplitude twice the distance found in part (a), and the kinetic coefficient of friction is μk=0.0850, what total distance does it travel before stopping? Assume it starts at the maximum amplitude.
Additional Problems
56.
-
Suppose you attach an object with mass m to a vertical spring originally at rest, and let it bounce up and down. You release the object from rest at the spring’s original rest length, the length of the spring in equilibrium, without the mass attached. The amplitude of the motion is the distance between the equilibrium position of the spring without the mass attached and the equilibrium position of the spring with the mass attached. (a) Show that the spring exerts an upward force of 2.00mg on the object at its lowest point. (b) If the spring has a force constant of 10.0 N/m, is hung horizontally, and the position of the free end of the spring is marked as y=0.00m, where is the new equilibrium position if a 0.25-kg-mass object is hung from the spring? (c) If the spring has a force constant of 10.0 M/m and a 0.25-kg-mass object is set in motion as described, find the amplitude of the oscillations. (d) Find the maximum velocity.
+
Suppose you attach an object with mass m to a vertical spring originally at rest, and let it bounce up and down. You release the object from rest at the spring’s original rest length, the length of the spring in equilibrium, without the mass attached. The amplitude of the motion is the distance between the equilibrium position of the spring without the mass attached and the equilibrium position of the spring with the mass attached. (a) Show that the spring exerts an upward force of 2.00mg on the object at its lowest point. (b) If the spring has a force constant of 10.0 N/m, is hung horizontally, and the position of the free end of the spring is marked as y=0.00m, where is the new equilibrium position if a 0.25-kg-mass object is hung from the spring? (c) If the spring has a force constant of 10.0 M/m and a 0.25-kg-mass object is set in motion as described, find the amplitude of the oscillations. (d) Find the maximum velocity.
@@ -2097,18 +2106,18 @@
Chapter Review
60.
-
A mass is placed on a frictionless, horizontal table. A spring (k=100N/m), which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=4.0cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t=3.00s.
+
A mass is placed on a frictionless, horizontal table. A spring (k=100N/m), which can be stretched or compressed, is placed on the table. A 5.00-kg mass is attached to one end of the spring, the other end is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=4.0cm and releases it from rest. The mass oscillates in SHM. (a) Determine the equations of motion. (b) Find the position, velocity, and acceleration of the mass at time t=3.00s.
61.
-
Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63m/s2.
+
Find the ratio of the new/old periods of a pendulum if the pendulum were transported from Earth to the Moon, where the acceleration due to gravity is 1.63m/s2.
62.
-
At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63m/s2, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.
+
At what rate will a pendulum clock run on the Moon, where the acceleration due to gravity is 1.63m/s2, if it keeps time accurately on Earth? That is, find the time (in hours) it takes the clock’s hour hand to make one revolution on the Moon.
@@ -2130,36 +2139,36 @@
Chapter Review
66.
-
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be modeled as a physical pendulum as a rod oscillating around one end. By what percentage will the period change if the temperature increases by 10°C? Assume the length of the rod changes linearly with temperature, where L=L0(1+αΔT) and the rod is made of brass (α=18×10−6°C−1).
+
Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be modeled as a physical pendulum as a rod oscillating around one end. By what percentage will the period change if the temperature increases by 10°C? Assume the length of the rod changes linearly with temperature, where L=L0(1+αΔT) and the rod is made of brass (α=18×10−6°C−1).
67.
-
A 2.00-kg block lies at rest on a frictionless table. A spring, with a spring constant of 100 N/m is attached to the wall and to the block. A second block of 0.50 kg is placed on top of the first block. The 2.00-kg block is gently pulled to a position x=+A and released from rest. There is a coefficient of friction of 0.45 between the two blocks. (a) What is the period of the oscillations? (b) What is the largest amplitude of motion that will allow the blocks to oscillate without the 0.50-kg block sliding off?
+
A 2.00-kg block lies at rest on a frictionless table. A spring, with a spring constant of 100 N/m is attached to the wall and to the block. A second block of 0.50 kg is placed on top of the first block. The 2.00-kg block is gently pulled to a position x=+A and released from rest. There is a coefficient of friction of 0.45 between the two blocks. (a) What is the period of the oscillations? (b) What is the largest amplitude of motion that will allow the blocks to oscillate without the 0.50-kg block sliding off?
Challenge Problems
68.
-
A suspension bridge oscillates with an effective force constant of 1.00×108N/m. (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1.00×104J of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude.
+
A suspension bridge oscillates with an effective force constant of 1.00×108N/m. (a) How much energy is needed to make it oscillate with an amplitude of 0.100 m? (b) If soldiers march across the bridge with a cadence equal to the bridge’s natural frequency and impart 1.00×104J of energy each second, how long does it take for the bridge’s oscillations to go from 0.100 m to 0.500 m amplitude.
69.
-
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00×105kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium?
+
Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.00×105kg on springs that have adjustable force constants. Its function is to dampen wind-driven oscillations of the building by oscillating at the same frequency as the building is being driven—the driving force is transferred to the object, which oscillates instead of the entire building. (a) What effective force constant should the springs have to make the object oscillate with a period of 2.00 s? (b) What energy is stored in the springs for a 2.00-m displacement from equilibrium?
70.
-
Parcels of air (small volumes of air) in a stable atmosphere (where the temperature increases with height) can oscillate up and down, due to the restoring force provided by the buoyancy of the air parcel. The frequency of the oscillations are a measure of the stability of the atmosphere. Assuming that the acceleration of an air parcel can be modeled as ∂2z′∂t2=gρo∂ρ(z)∂zz′, prove that z′=z0′et−N2√ is a solution, where N is known as the Brunt-Väisälä frequency. Note that in a stable atmosphere, the density decreases with height and parcel oscillates up and down.
+
Parcels of air (small volumes of air) in a stable atmosphere (where the temperature increases with height) can oscillate up and down, due to the restoring force provided by the buoyancy of the air parcel. The frequency of the oscillations are a measure of the stability of the atmosphere. Assuming that the acceleration of an air parcel can be modeled as ∂2z′∂t2=gρo∂ρ(z)∂zz′, prove that z′=z0′et−N2√ is a solution, where N is known as the Brunt-Väisälä frequency. Note that in a stable atmosphere, the density decreases with height and parcel oscillates up and down.
71.
-
Consider the van der Waals potential U(r)=Uo[(Ror)12−2(Ror)6], used to model the potential energy function of two molecules, where the minimum potential is at r=Ro. Find the force as a function of r. Consider a small displacement r=Ro+r′ and use the binomial theorem:
-
(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯,
+
Consider the van der Waals potential U(r)=Uo[(Ror)12−2(Ror)6], used to model the potential energy function of two molecules, where the minimum potential is at r=Ro. Find the force as a function of r. Consider a small displacement r=Ro+r′ and use the binomial theorem:
+
(1+x)n=1+nx+n(n−1)2!x2+n(n−1)(n−2)3!x3+⋯,
to show that the force does approximate a Hooke’s law force.
@@ -2171,7 +2180,7 @@
Chapter Review
73.
-
(a) The springs of a pickup truck act like a single spring with a force constant of 1.30×105N/m. By how much will the truck be depressed by its maximum load of 1000 kg? (b) If the pickup truck has four identical springs, what is the force constant of each?
+
(a) The springs of a pickup truck act like a single spring with a force constant of 1.30×105N/m. By how much will the truck be depressed by its maximum load of 1000 kg? (b) If the pickup truck has four identical springs, what is the force constant of each?
@@ -2357,7 +2366,7 @@ Types of Waves
A wave is a disturbance that propagates, or moves from the place it was created. There are three basic types of waves: mechanical waves, electromagnetic waves, and matter waves.
Basic mechanical waves are governed by Newton’s laws and require a medium. A medium is the substance a mechanical waves propagates through, and the medium produces an elastic restoring force when it is deformed. Mechanical waves transfer energy and momentum, without transferring mass. Some examples of mechanical waves are water waves, sound waves, and seismic waves. The medium for water waves is water; for sound waves, the medium is usually air. (Sound waves can travel in other media as well; we will look at that in more detail in Sound.) For surface water waves, the disturbance occurs on the surface of the water, perhaps created by a rock thrown into a pond or by a swimmer splashing the surface repeatedly. For sound waves, the disturbance is a change in air pressure, perhaps created by the oscillating cone inside a speaker or a vibrating tuning fork. In both cases, the disturbance is the oscillation of the molecules of the fluid. In mechanical waves, energy and momentum transfer with the motion of the wave, whereas the mass oscillates around an equilibrium point. (We discuss this in Energy and Power of a Wave.) Earthquakes generate seismic waves from several types of disturbances, including the disturbance of Earth’s surface and pressure disturbances under the surface. Seismic waves travel through the solids and liquids that form Earth. In this chapter, we focus on mechanical waves.
-Electromagnetic waves are associated with oscillations in electric and magnetic fields and do not require a medium. Examples include gamma rays, X-rays, ultraviolet waves, visible light, infrared waves, microwaves, and radio waves. Electromagnetic waves can travel through a vacuum at the speed of light, v=c=2.99792458×108m/s. For example, light from distant stars travels through the vacuum of space and reaches Earth. Electromagnetic waves have some characteristics that are similar to mechanical waves; they are covered in more detail in Electromagnetic Waves.
+Electromagnetic waves are associated with oscillations in electric and magnetic fields and do not require a medium. Examples include gamma rays, X-rays, ultraviolet waves, visible light, infrared waves, microwaves, and radio waves. Electromagnetic waves can travel through a vacuum at the speed of light, v=c=2.99792458×108m/s. For example, light from distant stars travels through the vacuum of space and reaches Earth. Electromagnetic waves have some characteristics that are similar to mechanical waves; they are covered in more detail in Electromagnetic Waves.
Matter waves are a central part of the branch of physics known as quantum mechanics. These waves are associated with protons, electrons, neutrons, and other fundamental particles found in nature. The theory that all types of matter have wave-like properties was first proposed by Louis de Broglie in 1924. Matter waves are discussed in Photons and Matter Waves.
@@ -2365,18 +2374,18 @@
Mechanical Waves
Mechanical waves exhibit characteristics common to all waves, such as amplitude, wavelength, period, frequency, and energy. All wave characteristics can be described by a small set of underlying principles.
-
The simplest mechanical waves repeat themselves for several cycles and are associated with simple harmonic motion. These simple harmonic waves can be modeled using some combination of sine and cosine functions. For example, consider the simplified surface water wave that moves across the surface of water as illustrated in Figure 16.3. Unlike complex ocean waves, in surface water waves, the medium, in this case water, moves vertically, oscillating up and down, whereas the disturbance of the wave moves horizontally through the medium. In Figure 16.3, the waves causes a seagull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The crest is the highest point of the wave, and the trough is the lowest part of the wave. The time for one complete oscillation of the up-and-down motion is the wave’s period T. The wave’s frequency is the number of waves that pass through a point per unit time and is equal to f=1/T. The period can be expressed using any convenient unit of time but is usually measured in seconds; frequency is usually measured in hertz (Hz), where 1Hz=1s−1.
-
The length of the wave is called the wavelength and is represented by the Greek letter lambda (λ), which is measured in any convenient unit of length, such as a centimeter or meter. The wavelength can be measured between any two similar points along the medium that have the same height and the same slope. In Figure 16.3, the wavelength is shown measured between two crests. As stated above, the period of the wave is equal to the time for one oscillation, but it is also equal to the time for one wavelength to pass through a point along the wave’s path.
-
The amplitude of the wave (A) is a measure of the maximum displacement of the medium from its equilibrium position. In the figure, the equilibrium position is indicated by the dotted line, which is the height of the water if there were no waves moving through it. In this case, the wave is symmetrical, the crest of the wave is a distance +A above the equilibrium position, and the trough is a distance −A below the equilibrium position. The units for the amplitude can be centimeters or meters, or any convenient unit of distance.
+
The simplest mechanical waves repeat themselves for several cycles and are associated with simple harmonic motion. These simple harmonic waves can be modeled using some combination of sine and cosine functions. For example, consider the simplified surface water wave that moves across the surface of water as illustrated in Figure 16.3. Unlike complex ocean waves, in surface water waves, the medium, in this case water, moves vertically, oscillating up and down, whereas the disturbance of the wave moves horizontally through the medium. In Figure 16.3, the waves causes a seagull to move up and down in simple harmonic motion as the wave crests and troughs (peaks and valleys) pass under the bird. The crest is the highest point of the wave, and the trough is the lowest part of the wave. The time for one complete oscillation of the up-and-down motion is the wave’s period T. The wave’s frequency is the number of waves that pass through a point per unit time and is equal to f=1/T. The period can be expressed using any convenient unit of time but is usually measured in seconds; frequency is usually measured in hertz (Hz), where 1Hz=1s−1.
+
The length of the wave is called the wavelength and is represented by the Greek letter lambda (λ), which is measured in any convenient unit of length, such as a centimeter or meter. The wavelength can be measured between any two similar points along the medium that have the same height and the same slope. In Figure 16.3, the wavelength is shown measured between two crests. As stated above, the period of the wave is equal to the time for one oscillation, but it is also equal to the time for one wavelength to pass through a point along the wave’s path.
+
The amplitude of the wave (A) is a measure of the maximum displacement of the medium from its equilibrium position. In the figure, the equilibrium position is indicated by the dotted line, which is the height of the water if there were no waves moving through it. In this case, the wave is symmetrical, the crest of the wave is a distance +A above the equilibrium position, and the trough is a distance −A below the equilibrium position. The units for the amplitude can be centimeters or meters, or any convenient unit of distance.
-
The water wave in the figure moves through the medium with a propagation velocity v⃗ . The magnitude of the wave velocity is the distance the wave travels in a given time, which is one wavelength in the time of one period, and the wave speed is the magnitude of wave velocity. In equation form, this is
+
Figure 16.3 An idealized surface water wave passes under a seagull that bobs up and down in simple harmonic motion. The wave has a wavelength λ, which is the distance between adjacent identical parts of the wave. The amplitude A of the wave is the maximum displacement of the wave from the equilibrium position, which is indicated by the dotted line. In this example, the medium moves up and down, whereas the disturbance of the surface propagates parallel to the surface at a speed v.
The water wave in the figure moves through the medium with a propagation velocity v⃗ . The magnitude of the wave velocity is the distance the wave travels in a given time, which is one wavelength in the time of one period, and the wave speed is the magnitude of wave velocity. In equation form, this is
This fundamental relationship holds for all types of waves. For water waves, v is the speed of a surface wave; for sound, v is the speed of sound; and for visible light, v is the speed of light.
@@ -2397,7 +2406,7 @@
-Figure 16.5 (a) This is a simple, graphical representation of a section of the stretched spring shown in Figure 16.4(b), representing the spring’s equilibrium position before any waves are induced on the spring. A point on the spring is marked by a blue dot. (b–g) Longitudinal waves are created by oscillating the end of the spring (not shown) back and forth along the x-axis. The longitudinal wave, with a wavelength λ, moves along the spring in the +x-direction with a wave speed v. For convenience, the wavelength is measured in (d). Note that the point on the spring that was marked with the blue dot moves back and forth a distance A from the equilibrium position, oscillating around the equilibrium position of the point.
+Figure 16.5 (a) This is a simple, graphical representation of a section of the stretched spring shown in Figure 16.4(b), representing the spring’s equilibrium position before any waves are induced on the spring. A point on the spring is marked by a blue dot. (b–g) Longitudinal waves are created by oscillating the end of the spring (not shown) back and forth along the x-axis. The longitudinal wave, with a wavelength λ, moves along the spring in the +x-direction with a wave speed v. For convenience, the wavelength is measured in (d). Note that the point on the spring that was marked with the blue dot moves back and forth a distance A from the equilibrium position, oscillating around the equilibrium position of the point.
Waves may be transverse, longitudinal, or a combination of the two. Examples of transverse waves are the waves on stringed instruments or surface waves on water, such as ripples moving on a pond. Sound waves in air and water are longitudinal. With sound waves, the disturbances are periodic variations in pressure that are transmitted in fluids. Fluids do not have appreciable shear strength, and for this reason, the sound waves in them are longitudinal waves. Sound in solids can have both longitudinal and transverse components, such as those in a seismic wave. Earthquakes generate seismic waves under Earth’s surface with both longitudinal and transverse components (called compressional or P-waves and shear or S-waves, respectively). The components of seismic waves have important individual characteristics—they propagate at different speeds, for example. Earthquakes also have surface waves that are similar to surface waves on water. Ocean waves also have both transverse and longitudinal components.
Example 16.1
Wave on a String
@@ -2406,16 +2415,16 @@
The speed of the wave can be derived by dividing the distance traveled by the time.
The period of the wave is the inverse of the frequency of the driving force.
-
The wavelength can be found from the speed and the period v=λ/T.
+
The wavelength can be found from the speed and the period v=λ/T.
Solution
- The first wave traveled 30.00 m in 6.00 s:
-
+
- The period is equal to the inverse of the frequency:
-
+
- The wavelength is equal to the velocity times the period:
-λ=vT=5.00ms(0.50s)=2.50m.
+λ=vT=5.00ms(0.50s)=2.50m.
Significance
The frequency of the wave produced by an oscillating driving force is equal to the frequency of the driving force.
@@ -2430,7 +2439,7 @@
Example 16.2
Characteristics of a Wave
-A transverse mechanical wave propagates in the positive x-direction through a spring (as shown in Figure 16.4(a)) with a constant wave speed, and the medium oscillates between +A and −A around an equilibrium position. The graph in Figure 16.6 shows the height of the spring (y) versus the position (x), where the x-axis points in the direction of propagation. The figure shows the height of the spring versus the x-position at t=0.00s as a dotted line and the wave at t=3.00s as a solid line. (a) Determine the wavelength and amplitude of the wave. (b) Find the propagation velocity of the wave. (c) Calculate the period and frequency of the wave.
+A transverse mechanical wave propagates in the positive
x-direction through a spring (as shown in
Figure 16.4(a)) with a constant wave speed, and the medium oscillates between
+A and
−A around an equilibrium position. The graph in
Figure 16.6 shows the height of the spring (
y) versus the position (
x), where the
x-axis points in the direction of propagation. The figure shows the height of the spring versus the
x-position at
t=0.00s as a dotted line and the wave at
t=3.00s as a solid line. (a) Determine the wavelength and amplitude of the wave. (b) Find the propagation velocity of the wave. (c) Calculate the period and frequency of the wave.
-
In this chapter, we consider only string with a constant linear density. If the linear density is constant, then the mass (Δm) of a small length of string (Δx) is Δm=μΔx. For example, if the string has a length of 2.00 m and a mass of 0.06 kg, then the linear density is μ=0.06kg2.00m=0.03kgm. If a 1.00-mm section is cut from the string, the mass of the 1.00-mm length is Δm=μΔx=(0.03kgm)0.001m=3.00×10−5kg. The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed.
+
In this chapter, we consider only string with a constant linear density. If the linear density is constant, then the mass (Δm) of a small length of string (Δx) is Δm=μΔx. For example, if the string has a length of 2.00 m and a mass of 0.06 kg, then the linear density is μ=0.06kg2.00m=0.03kgm. If a 1.00-mm section is cut from the string, the mass of the 1.00-mm length is Δm=μΔx=(0.03kgm)0.001m=3.00×10−5kg. The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed.
Wave Speed on a String under Tension
-To see how the speed of a wave on a string depends on the tension and the linear density, consider a pulse sent down a taut string (Figure 16.13). When the taut string is at rest at the equilibrium position, the tension in the string FT is constant. Consider a small element of the string with a mass equal to Δm=μΔx. The mass element is at rest and in equilibrium and the force of tension of either side of the mass element is equal and opposite.
+To see how the speed of a wave on a string depends on the tension and the linear density, consider a pulse sent down a taut string (Figure 16.13). When the taut string is at rest at the equilibrium position, the tension in the string FT is constant. Consider a small element of the string with a mass equal to Δm=μΔx. The mass element is at rest and in equilibrium and the force of tension of either side of the mass element is equal and opposite.