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Contents

Section 17.1 - Period, frequency, and amplitude

Section 17.2 - Simple harmonic motion

Section 17.3 - Proofs (optional)

Section 17.4 - Summary

Section 17.1 - Period, frequency, and amplitude

Section 17.2 - Simple harmonic motion

Section 17.3 - Proofs (optional)

Section 17.4 - Summary

Dandelion. Cello. Read those two words, and your brain instantly conjures a stream of associations, the most prominent of which have to do with vibrations. Our mental category of “dandelion-ness” is strongly linked to the color of light waves that vibrate about half a million billion times a second: yellow. The velvety throb of a cello has as its most obvious characteristic a relatively low musical pitch --- the note you are spontaneously imagining right now might be one whose sound vibrations repeat at a rate of a hundred times a second.

Evolution has designed our two most important senses around the assumption that not only will our environment be drenched with information-bearing vibrations, but in addition those vibrations will often be repetitive, so that we can judge colors and pitches by the rate of repetition. Granting that we do sometimes encounter non-repeating waves such as the consonant “sh,” which has no recognizable pitch, why was Nature's assumption of repetition nevertheless so right in general?

Repeating phenomena occur throughout nature, from the orbits of electrons in atoms to the reappearance of Halley's Comet every 75 years. Ancient cultures tended to attribute repetitious phenomena like the seasons to the cyclical nature of time itself, but we now have a less mystical explanation. Suppose that instead of Halley's Comet's true, repeating elliptical orbit that closes seamlessly upon itself with each revolution, we decide to take a pen and draw a whimsical alternative path that never repeats. We will not be able to draw for very long without having the path cross itself. But at such a crossing point, the comet has returned to a place it visited once before, and since its potential energy is the same as it was on the last visit, conservation of energy proves that it must again have the same kinetic energy and therefore the same speed. Not only that, but the comet's direction of motion cannot be randomly chosen, because angular momentum must be conserved as well. Although this falls short of being an ironclad proof that the comet's orbit must repeat, it no longer seems surprising that it does.

Conservation laws, then, provide us with a good reason why repetitive motion is so prevalent in the universe. But it goes deeper than that. Up to this point in your study of physics, I have been indoctrinating you with a mechanistic vision of the universe as a giant piece of clockwork. Breaking the clockwork down into smaller and smaller bits, we end up at the atomic level, where the electrons circling the nucleus resemble --- well, little clocks! From this point of view, particles of matter are the fundamental building blocks of everything, and vibrations and waves are just a couple of the tricks that groups of particles can do. But at the beginning of the 20th century, the tables were turned. A chain of discoveries initiated by Albert Einstein led to the realization that the so-called subatomic “particles” were in fact waves. In this new world-view, it is vibrations and waves that are fundamental, and the formation of matter is just one of the tricks that waves can do.

Figure b shows our most basic example of a vibration. With no forces on it, the spring assumes its equilibrium length, b/1. It can be stretched, 2, or compressed, 3. We attach the spring to a wall on the left and to a mass on the right. If we now hit the mass with a hammer, 4, it oscillates as shown in the series of snapshots, 4-13. If we assume that the mass slides back and forth without friction and that the motion is one-dimensional, then conservation of energy proves that the motion must be repetitive. When the block comes back to its initial position again, 7, its potential energy is the same again, so it must have the same kinetic energy again. The motion is in the opposite direction, however. Finally, at 10, it returns to its initial position with the same kinetic energy and the same direction of motion. The motion has gone through one complete cycle, and will now repeat forever in the absence of friction.

The usual physics terminology for motion that repeats itself over and over is periodic motion, and the time required for one repetition is called the period, \(T\). (The symbol \(P\) is not used because of the possible confusion with momentum.) One complete repetition of the motion is called a cycle.

We are used to referring to short-period sound vibrations as “high” in pitch, and it sounds odd to have to say that high pitches have low periods. It is therefore more common to discuss the rapidity of a vibration in terms of the number of vibrations per second, a quantity called the frequency, \(f\). Since the period is the number of seconds per cycle and the frequency is the number of cycles per second, they are reciprocals of each other,

\[\begin{equation*}
f = 1/T .
\end{equation*}\]

A Victorian parlor trick was to listen to the pitch of a fly's buzz, reproduce the musical note on the piano, and announce how many times the fly's wings had flapped in one second. If the fly's wings flap, say, 200 times in one second, then the frequency of their motion is \(f=200/1\ \text{s}=200\ \text{s}^{-1}\). The period is one 200th of a second, \(T=1/f=(1/200)\ \text{s}=0.005\ \text{s}\).

Units of inverse second, \(\text{s}^{-1}\), are awkward in speech, so an abbreviation has been created. One Hertz, named in honor of a pioneer of radio technology, is one cycle per second. In abbreviated form, \(1\ \text{Hz}=1\ \text{s}^{-1}\). This is the familiar unit used for the frequencies on the radio dial.

\(\triangleright\) KKJZ's frequency is 88.1 MHz. What does this mean, and what period does this correspond to?

\(\triangleright\) The metric prefix M- is mega-, i.e., millions. The radio waves emitted by KKJZ's transmitting antenna vibrate 88.1 million times per second. This corresponds to a period of

\[\begin{equation*}
T = 1/f= 1.14\times10^{-8}\ \text{s} .
\end{equation*}\]

This example shows a second reason why we normally speak in terms of frequency rather than period: it would be painful to have to refer to such small time intervals routinely. I could abbreviate by telling people that KKJZ's period was 11.4 nanoseconds, but most people are more familiar with the big metric prefixes than with the small ones.

Units of frequency are also commonly used to specify the speeds of computers. The idea is that all the little circuits on a computer chip are synchronized by the very fast ticks of an electronic clock, so that the circuits can all cooperate on a task without getting ahead or behind. Adding two numbers might require, say, 30 clock cycles. Microcomputers these days operate at clock frequencies of about a gigahertz.

We have discussed how to measure how fast something vibrates, but not how big the vibrations are. The general term for this is amplitude, \(A\). The definition of amplitude depends on the system being discussed, and two people discussing the same system may not even use the same definition. In the example of the block on the end of the spring, e/1, the amplitude will be measured in distance units such as cm. One could work in terms of the distance traveled by the block from the extreme left to the extreme right, but it would be somewhat more common in physics to use the distance from the center to one extreme. The former is usually referred to as the peak-to-peak amplitude, since the extremes of the motion look like mountain peaks or upside-down mountain peaks on a graph of position versus time.

In other situations we would not even use the same units for amplitude. The amplitude of a child on a swing, or a pendulum, e/2, would most conveniently be measured as an angle, not a distance, since her feet will move a greater distance than her head. The electrical vibrations in a radio receiver would be measured in electrical units such as volts or amperes.

If we actually construct the mass-on-a-spring system discussed in the previous section and measure its motion accurately, we will find that its \(x-t\) graph is nearly a perfect sine-wave shape, as shown in figure f/1. (We call it a “sine wave” or “sinusoidal” even if it is a cosine, or a sine or cosine shifted by some arbitrary horizontal amount.) It may not be surprising that it is a wiggle of this general sort, but why is it a specific mathematically perfect shape? Why is it not a sawtooth shape like 2 or some other shape like 3? The mystery deepens as we find that a vast number of apparently unrelated vibrating systems show the same mathematical feature. A tuning fork, a sapling pulled to one side and released, a car bouncing on its shock absorbers, all these systems will exhibit sine-wave motion under one condition: the amplitude of the motion must be small.

It is not hard to see intuitively why extremes of amplitude would act differently. For example, a car that is bouncing lightly on its shock absorbers may behave smoothly, but if we try to double the amplitude of the vibrations the bottom of the car may begin hitting the ground, f/4. (Although we are assuming for simplicity in this chapter that energy is never dissipated, this is clearly not a very realistic assumption in this example. Each time the car hits the ground it will convert quite a bit of its potential and kinetic energy into heat and sound, so the vibrations would actually die out quite quickly, rather than repeating for many cycles as shown in the figure.)

The key to understanding how an object vibrates is to know how the force on the object depends on the object's position. If an object is vibrating to the right and left, then it must have a leftward force on it when it is on the right side, and a rightward force when it is on the left side. In one dimension, we can represent the direction of the force using a positive or negative sign, and since the force changes from positive to negative there must be a point in the middle where the force is zero. This is the equilibrium point, where the object would stay at rest if it was released at rest. For convenience of notation throughout this chapter, we will define the origin of our coordinate system so that \(x\) equals zero at equilibrium.

The simplest example is the mass on a spring, for which the force on the mass is given by Hooke's law,

\[\begin{equation*}
F = -kx .
\end{equation*}\]

We can visualize the behavior of this force using a graph of \(F\) versus \(x\), as shown in figure g. The graph is a line, and the spring constant, \(k\), is equal to minus its slope. A stiffer spring has a larger value of \(k\) and a steeper slope. Hooke's law is only an approximation, but it works very well for most springs in real life, as long as the spring isn't compressed or stretched so much that it is permanently bent or damaged.

The following important theorem, whose proof is given in optional section 17.3, relates the motion graph to the force graph.

{Theorem: A linear force graph makes a sinusoidal motion graph.}

If the total force on a vibrating object depends only on the object's position, and is related to the objects displacement from equilibrium by an equation of the form \(F=-kx\), then the object's motion displays a sinusoidal graph with period \(T=2\pi\sqrt{m/k}\).

Even if you do not read the proof, it is not too hard to understand why the equation for the period makes sense. A greater mass causes a greater period, since the force will not be able to whip a massive object back and forth very rapidly. A larger value of \(k\) causes a shorter period, because a stronger force can whip the object back and forth more rapidly.

This may seem like only an obscure theorem about the mass-on-a-spring system, but figure h shows it to be far more general than that. Figure h/1 depicts a force curve that is not a straight line. A system with this \(F-x\) curve would have large-amplitude vibrations that were complex and not sinusoidal. But the same system would exhibit sinusoidal small-amplitude vibrations. This is because any curve looks linear from very close up. If we magnify the \(F-x\) graph as shown in figure h/2, it becomes very difficult to tell that the graph is not a straight line. If the vibrations were confined to the region shown in h/2, they would be very nearly sinusoidal. This is the reason why sinusoidal vibrations are a universal feature of all vibrating systems, if we restrict ourselves to small amplitudes. The theorem is therefore of great general significance. It applies throughout the universe, to objects ranging from vibrating stars to vibrating nuclei. A sinusoidal vibration is known as simple harmonic motion.

Until now we have not even mentioned the most counterintuitive aspect of the equation \(T=2\pi\sqrt{m/k}\): it does not depend on amplitude at all. Intuitively, most people would expect the mass-on-a-spring system to take longer to complete a cycle if the amplitude was larger. (We are comparing amplitudes that are different from each other, but both small enough that the theorem applies.) In fact the larger-amplitude vibrations take the same amount of time as the small-amplitude ones. This is because at large amplitudes, the force is greater, and therefore accelerates the object to higher speeds.

Legend has it that this fact was first noticed by Galileo during what was apparently a less than enthralling church service. A gust of wind would now and then start one of the chandeliers in the cathedral swaying back and forth, and he noticed that regardless of the amplitude of the vibrations, the period of oscillation seemed to be the same. Up until that time, he had been carrying out his physics experiments with such crude time-measuring techniques as feeling his own pulse or singing a tune to keep a musical beat. But after going home and testing a pendulum, he convinced himself that he had found a superior method of measuring time. Even without a fancy system of pulleys to keep the pendulum's vibrations from dying down, he could get very accurate time measurements, because the gradual decrease in amplitude due to friction would have no effect on the pendulum's period. (Galileo never produced a modern-style pendulum clock with pulleys, a minute hand, and a second hand, but within a generation the device had taken on the form that persisted for hundreds of years after.)

\(\triangleright\) Compare the periods of pendula having bobs with different masses.

\(\triangleright\) From the equation \(T=2\pi\sqrt{m/k}\), we might expect that a larger mass would lead to a longer period. However, increasing the mass also increases the forces that act on the pendulum: gravity and the tension in the string. This increases \(k\) as well as \(m\), so the period of a pendulum is independent of \(m\).

◊

Suppose that a pendulum has a rigid arm mounted on a bearing, rather than a string tied at its top with a knot. The bob can then oscillate with center-to-side amplitudes greater than \(90°\). For the maximum amplitude of \(180°\), what can you say about the period?

◊

In the language of calculus, Newton's second law for a simple harmonic oscillator can be written in the form \(d^2 x/dt^2=-(...)x\), where “...” refers to a constant, and the minus sign says that if we pull the object away from equilibrium, a restoring force tries to bring it back to equlibrium, which is the opposite direction. This is why we get motion that looks like a sine or cosine function: these are functions that, when differentiated twice, give back the original function but with an opposite sign. Now consider the example described in discussion question A, where a pendulum is upright or nearly upright. How does the analysis play out differently?

In this section we prove (1) that a linear \(F-x\) graph gives sinusoidal motion, (2) that the period of the motion is \(2\pi\sqrt{m/k}\), and (3) that the period is independent of the amplitude. You may omit this section without losing the continuity of the chapter.

The basic idea of the proof can be understood by imagining that you are watching a child on a merry-go-round from far away. Because you are in the same horizontal plane as her motion, she appears to be moving from side to side along a line. Circular motion viewed edge-on doesn't just look like any kind of back-and-forth motion, it looks like motion with a sinusoidal \(x-t\) graph, because the sine and cosine functions can be defined as the \(x\) and \(y\) coordinates of a point at angle \(\theta \) on the unit circle. The idea of the proof, then, is to show that an object acted on by a force that varies as \(F=-kx\) has motion that is identical to circular motion projected down to one dimension. The \(v^2/r\) expression will also fall out at the end.

For an object performing uniform circular motion, we have

\[\begin{equation*}
|\mathbf{a}| = \frac{v^2}{r} .
\end{equation*}\]

The \(x\) component of the acceleration is therefore

\[\begin{equation*}
a_x = \frac{v^2}{r}\cos\theta ,
\end{equation*}\]

where \(\theta \) is the angle measured counterclockwise from the \(x\) axis. Applying Newton's second law,

\[\begin{align*}
\frac{F_x}{m} &= -\frac{v^2}{r}\cos\theta , \text{so} \\
F_x &= -m\frac{v^2}{r}\cos\theta .
\end{align*}\]

Since our goal is an equation involving the period, it is natural to eliminate the variable \(v = \text{circumference}/T=2\pi r/T\), giving

\[\begin{equation*}
F_x = -\frac{4\pi^2 mr}{T^2}\cos\theta .
\end{equation*}\]

The quantity \(r \cos \theta \) is the same as \(x\), so we have

\[\begin{equation*}
F_x = -\frac{4\pi^2 m}{T^2}\: x .
\end{equation*}\]

Since everything is constant in this equation except for \(x\), we have proved that motion with force proportional to \(x\) is the same as circular motion projected onto a line, and therefore that a force proportional to \(x\) gives sinusoidal motion. Finally, we identify the constant factor of \(4\pi^2m/T^2\) with \(k\), and solving for \(T\) gives the desired equation for the period,

\[\begin{equation*}
T = 2\pi\sqrt{\frac{m}{k}} .
\end{equation*}\]

Since this equation is independent of \(r\), \(T\) is independent of the amplitude, subject to the initial assumption of perfect \(F=-kx\) behavior, which in reality will only hold approximately for small \(x\).

*periodic motion* — motion that repeats itself over and over

*period* — the time required for one cycle of a periodic motion

*frequency* — the number of cycles per second, the inverse of the period

*amplitude* — the amount of vibration, often measured from the
center to one side; may have different units depending on
the nature of the vibration

*simple harmonic motion* — motion whose \(x-t\) graph is a sine wave

\(T\) — period

\(f\) — frequency

\(A\) — amplitude

\(k\) — the slope of the graph of \(F\) versus \(x\), where \(F\) is the total force acting on an object and \(x\) is the object's position; for a spring, this is known as the spring constant.

\(\nu\) — The Greek letter \(\nu \), nu, is used in many books for frequency.

\(\omega\) — The Greek letter \(\omega \), omega, is often used as an abbreviation for \(2\pi f\).

{}

Periodic motion is common in the world around us because of conservation laws. An important example is one-dimensional motion in which the only two forms of energy involved are potential and kinetic; in such a situation, conservation of energy requires that an object repeat its motion, because otherwise when it came back to the same point, it would have to have a different kinetic energy and therefore a different total energy.

Not only are periodic vibrations very common, but small-amplitude vibrations are always sinusoidal as well. That is, the \(x-t\) graph is a sine wave. This is because the graph of force versus position will always look like a straight line on a sufficiently small scale. This type of vibration is called simple harmonic motion. In simple harmonic motion, the period is independent of the amplitude, and is given by

\[\begin{equation*}
T=2\pi\sqrt{m/k} .
\end{equation*}\]

\begin{homeworkforcelabel}{shmfreq}{1}{}{1}Find an equation for the frequency of simple harmonic motion in terms of \(k\) and \(m\). (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{sperm}{1}{}{2}Many single-celled organisms propel themselves through water with long tails, which they wiggle back and forth. (The most obvious example is the sperm cell.) The frequency of the tail's vibration is typically about 10-15 Hz. To what range of periods does this range of frequencies correspond? \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pendulum}{1}{}{3}(a) Pendulum 2 has a string twice as long as pendulum 1. If we define \(x\) as the distance traveled by the bob along a circle away from the bottom, how does the \(k\) of pendulum 2 compare with the \(k\) of pendulum 1? Give a numerical ratio. [Hint: the total force on the bob is the same if the angles away from the bottom are the same, but equal angles do not correspond to equal values of \(x\).]

(b) Based on your answer from part (a), how does the period of pendulum 2 compare with the period of pendulum 1? Give a numerical ratio. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{piston}{1}{}{4}A pneumatic spring consists of a piston riding on top of the air in a cylinder. The upward force of the air on the piston is given by \(F_{air}=ax^{-1.4}\), where \(a\) is a constant with funny units of \(\text{N}\!\cdot\!\text{m}^{1.4}\). For simplicity, assume the air only supports the weight, \(F_W\), of the piston itself, although in practice this device is used to support some other object. The equilibrium position, \(x_0\), is where \(F_W\) equals \(-F_{air}\). (Note that in the main text I have assumed the equilibrium position to be at \(x=0\), but that is not the natural choice here.) Assume friction is negligible, and consider a case where the amplitude of the vibrations is very small. Let \(a=1.0\ \text{N}\!\cdot\!\text{m}^{1.4}\), \(x_0=1.00\ \text{m}\), and \(F_W=-1.00\ \text{N}\). The piston is released from \(x=1.01\ \text{m}\). Draw a neat, accurate graph of the total force, \(F\), as a function of \(x\), on graph paper, covering the range from \(x=0.98\ \text{m}\) to 1.02 \(\text{m}\). Over this small range, you will find that the force is very nearly proportional to \(x-x_0\). Approximate the curve with a straight line, find its slope, and derive the approximate period of oscillation. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{pistonxt}{1}{}{5}Consider the same pneumatic piston described in problem 4, but now imagine that the oscillations are not small. Sketch a graph of the total force on the piston as it would appear over this wider range of motion. For a wider range of motion, explain why the vibration of the piston about equilibrium is not simple harmonic motion, and sketch a graph of \(x\) vs \(t\), showing roughly how the curve is different from a sine wave. [Hint: Acceleration corresponds to the curvature of the \(x-t\) graph, so if the force is greater, the graph should curve around more quickly.] \end{homeworkforcelabel}

\begin{homeworkforcelabel}{bobbing}{1}{}{6}Archimedes' principle states that an object partly or wholly immersed in fluid experiences a buoyant force equal to the weight of the fluid it displaces. For instance, if a boat is floating in water, the upward pressure of the water (vector sum of all the forces of the water pressing inward and upward on every square inch of its hull) must be equal to the weight of the water displaced, because if the boat was instantly removed and the hole in the water filled back in, the force of the surrounding water would be just the right amount to hold up this new “chunk” of water. (a) Show that a cube of mass \(m\) with edges of length \(b\) floating upright (not tilted) in a fluid of density \(\rho \) will have a draft (depth to which it sinks below the waterline) \(h\) given at equilibrium by \(h_0=m/b^2\rho\). (b) Find the total force on the cube when its draft is \(h\), and verify that plugging in \(h-h_0\) gives a total force of zero. (c) Find the cube's period of oscillation as it bobs up and down in the water, and show that can be expressed in terms of and \(g\) only. (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{codornices}{2}{}{7}The figure shows a see-saw with two springs at Codornices Park in Berkeley, California. Each spring has spring constant \(k\), and a kid of mass \(m\) sits on each seat. (a) Find the period of vibration in terms of the variables \(k\), \(m\), \(a\), and \(b\). (b) Discuss the special case where \(a=b\), rather than \(a>b\) as in the real see-saw. (c) Show that your answer to part a also makes sense in the case of \(b=0\). (answer check available at lightandmatter.com) \end{homeworkforcelabel}

\begin{homeworkforcelabel}{shm-units}{1}{}{8}Show that the equation \(T=2\pi\sqrt{m/k}\) has units that make sense. \end{homeworkforcelabel}

\begin{homeworkforcelabel}{oblate-earth}{1}{}{9}A hot scientific question of the 18th century was the shape of the earth: whether its radius was greater at the equator than at the poles, or the other way around. One method used to attack this question was to measure gravity accurately in different locations on the earth using pendula. If the highest and lowest latitudes accessible to explorers were 0 and 70 degrees, then the the strength of gravity would in reality be observed to vary over a range from about 9.780 to 9.826 \(\text{m}/\text{s}^2\). This change, amounting to 0.046 \(\text{m}/\text{s}^2\), is greater than the 0.022 \(\text{m}/\text{s}^2\) effect to be expected if the earth had been spherical. The greater effect occurs because the equator feels a reduction due not just to the acceleration of the spinning earth out from under it, but also to the greater radius of the earth at the equator. What is the accuracy with which the period of a one-second pendulum would have to be measured in order to prove that the earth was not a sphere, and that it bulged at the equator? \end{homeworkforcelabel}

\begin{handson}{}{Vibrations}{\onecolumn} Equipment:

- air track and carts of two different masses
- springs
- spring scales

\inlinefignocaptionnoresize{../../../share/mechanics/figs/ex-vibrations}

Place the cart on the air track and attach springs so that it can vibrate.

1. Test whether the period of vibration depends on amplitude. Try at least one moderate amplitude, for which the springs do not go slack, at least one amplitude that is large enough so that they do go slack, and one amplitude that's the very smallest you can possibly observe.

2. Try a cart with a different mass. Does the period change by the expected factor, based on the equation \(T=2\pi\sqrt{m/k}\)?

3. Use a spring scale to pull the cart away from equilibrium, and make a graph of force versus position. Is it linear? If so, what is its slope?

4. Test the equation \(T=2\pi\sqrt{m/k}\) numerically. \end{handson}

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.