Precalculus

Simple harmonic motion

A weight on a spring, pulled down and let go, bobs up and down on its own. Where will it be a second later, how long does each bounce take, and where along its path is it moving fastest?

Where you are 3 in Modeling things that repeat

Trigonometry7

Trigonometric functions and their graphs8

Modeling things that repeat3

Height on a turning wheel

Simple harmonic motion

Tides, daylight and sound

Why this exists

Hang a weight on a spring and it settles at a resting place, its equilibrium. Pull it down and let go, and it bobs up and down past that place again and again. The tip of a diving board after a jump does the same, and so does a car body riding on its springs after a bump. In each case the spring pulls the object back toward its resting place, harder the farther away it is; OpenStax describes this restoring force as directly proportional to the displacement.

Such motion repeats on a steady cycle, so a sine or cosine of time can describe it, and questions about the object become questions about a function: where it is at a given moment, how long a bounce lasts, and where it moves fastest. This lesson builds on Amplitude and reflection, Finding the period from the equation and Writing an equation from a wave. Its neighbors Height on a turning wheel and Tides, daylight and sound use the same functions for a turning wheel and a harbor.

Definition

Measure the object’s position by its displacement d from equilibrium, positive on one side (here, up) and negative on the other, and let t be time in seconds. The object is in simple harmonic motion when

d(t) = a\cos(\omega t) \quad \text{or} \quad d(t) = a\sin(\omega t)

for constants a \ne 0 and \omega > 0.

|a| is the amplitude, the greatest distance from equilibrium.
\omega (omega) is the angular frequency, in radians per second.
The period P is the time for one full cycle, out and back.
The frequency f is the number of cycles per second, measured in hertz (\text{Hz}).

The cosine form starts at an extreme: at t = 0 it gives d = a. It fits an object pulled to d = a and let go. The sine form starts at equilibrium, d(0) = 0, moving up when a > 0 and down when a < 0. It fits an object struck or pushed while at rest in its resting place.

Intuition

Figure 1Jeruto’s weight over two bounces. It turns at the dots on d = 15 and d = -15, and passes the rest position d = 0 twice in every 0.8 seconds.

The graph shows Jeruto’s weight, pulled 15 cm below rest and let go. It climbs, slows, stops for an instant at the top, comes back down, and stops again at the bottom. These two places, d = |a| and d = -|a|, are its turning points. Between them it is fastest in the middle of its path, where the graph is steepest.

There is a second picture of the same motion. Picture a point going around a circle of radius |a| at a steady rate, \omega radians each second, and watch only its height. That height is a sine or cosine of \omega t, the same function as the weight’s displacement. When the point is level with the center, it is moving straight up or down, so its height changes at its full speed. At the top and bottom of the circle it is moving sideways, so its height stops changing for an instant.

Figure 2A point going around a circle at a steady rate casts a shadow on the vertical line. The shadow moves exactly like the weight, fastest as it passes the center and still for an instant at the top and bottom.

Results

Period, frequency, angular frequency. For d(t) = a\cos(\omega t) or d(t) = a\sin(\omega t),

P = \frac{2\pi}{\omega}, \qquad f = \frac{1}{P} = \frac{\omega}{2\pi}, \qquad \omega = 2\pi f

Where it turns. The object turns at d = |a| and d = -|a|, once each per cycle, and passes equilibrium twice per cycle, once in each direction. The amplitude does not appear in P: in the model, a bigger pull changes how far the object swings, not how long a swing takes.

Where it is fastest. The speed is greatest, |a|\omega, each time the object passes equilibrium, and it is zero for an instant at each turning point.

Damping. When the swings shrink by the same factor r each second, with 0 < r < 1,

d(t) = a\,r^t\cos(\omega t) \quad \text{or} \quad d(t) = a\,r^t\sin(\omega t)

This is damped harmonic motion. The factor a\,r^t is an exponential decay; writing r = e^{-c} gives OpenStax’s form a e^{-ct}\cos(\omega t) with damping constant c = -\ln r.

Why it is true

The period. The cosine and sine repeat when their input grows by 2\pi. The input \omega t grows by 2\pi exactly when t grows by \displaystyle \frac{2\pi}{\omega}, so \displaystyle P = \frac{2\pi}{\omega}. A cycle that lasts P seconds happens \displaystyle \frac{1}{P} times each second, so \displaystyle f = \frac{1}{P} = \frac{\omega}{2\pi}, and solving for \omega gives \omega = 2\pi f. In words, one cycle is 2\pi radians of input, so f cycles each second is 2\pi f radians each second.

The turning points. \cos(\omega t) takes each of the values 1 and -1 once per cycle and the value 0 twice, so d reaches \pm|a| once each and 0 twice per cycle. The same holds for sine. Between a -|a| and the next |a| the function only rises, and between an |a| and the next -|a| it only falls, so the object reverses direction only at those two values.

The speed. Take d(t) = a\sin(\omega t) with a > 0 (the other forms are this one shifted in time or flipped). Let a point Q go around a circle of radius a, centered on the rest position, turning \omega radians each second. By Arc length and sector area in radians, it covers a\omega units of arc each second, so its speed is a\omega (Angular and linear speed). After t seconds it has turned through the angle \omega t from the level of the center, so its height above the center is a\sin(\omega t) = d(t). The object’s motion is the up and down part of Q’s motion. As Q passes the level of the center, its path points straight up or down, so all of its speed a\omega goes into changing its height. Near the top or bottom its path points sideways, and its height hardly changes: the object stops there for an instant. Everywhere between, part of Q’s motion is sideways, so the object moves more slowly than a\omega.

That a spring follows a sine or cosine at all comes from physics and calculus; here the model is taken as given.

Worked examples

Example 1

Jeruto’s spring

Jeruto pulls a weight on a spring 15 cm below its rest position and lets go at t = 0. It makes one full bounce every 0.8 seconds. With up positive, write a model for its displacement, give the amplitude, frequency and angular frequency, and find where it is at t = 0.2, 0.4 and 0.8.

Solution

The weight starts at an extreme, so the cosine form fits. At t = 0 it is 15 cm below rest, so d(0) = -15 and a = -15. The period is P = 0.8, so

\omega = \frac{2\pi}{0.8} = \frac{5\pi}{2}, \qquad f = \frac{1}{0.8} = 1.25

The model is

d(t) = -15\cos\left(\frac{5\pi}{2}t\right)

The amplitude is |-15| = 15 cm, the frequency 1.25 hertz, and \displaystyle \omega = \frac{5\pi}{2} radians per second. At t = 0.2 the input is \displaystyle \frac{\pi}{2} and d = -15 \cdot 0 = 0: the weight is passing its rest position, a quarter period after release. At t = 0.4 the input is \pi and d = -15(-1) = 15, the top turning point. At t = 0.8 the input is 2\pi and d = -15: one full bounce, back where it started.

Example 2

Chebet’s diving board

After a jump, the tip of Chebet’s diving board moves with displacement d(t) = 6\sin(3\pi t) cm, up positive. Describe how it starts, give its period and frequency, and find every time in the first 2 seconds when the tip is at its highest and at its lowest.

Solution

This is the sine form with a = 6 > 0, so the tip starts at rest position moving up, and it swings 6 cm each way. Here \omega = 3\pi, so \displaystyle P = \frac{2\pi}{3\pi} = \frac{2}{3} second and \displaystyle f = \frac{3}{2} = 1.5 hertz.

The tip is highest when \sin(3\pi t) = 1, that is when \displaystyle 3\pi t = \frac{\pi}{2} + 2\pi n for n any integer. Dividing by 3\pi gives \displaystyle t = \frac{1}{6} + \frac{2n}{3}, and the times in [0, 2] are \displaystyle t = \frac{1}{6}, \frac{5}{6}, \frac{3}{2}. It is lowest when \displaystyle 3\pi t = \frac{3\pi}{2} + 2\pi n, so \displaystyle t = \frac{1}{2} + \frac{2n}{3}: \displaystyle t = \frac{1}{2}, \frac{7}{6}, \frac{11}{6}. Each low comes half a period, \displaystyle \frac{1}{3} second, after a high, as it should.

Example 3

From a frequency

Jeruto hangs a lighter weight on the spring, lifts it 6 cm above rest and lets go. It makes 2 full cycles each second. Write a model, find the first time it is 3 cm below rest, and find where it is at t = 0.3.

Solution

The release is at an extreme, 6 cm up, so a = 6 and the cosine form fits. The frequency is f = 2, so \omega = 2\pi \cdot 2 = 4\pi and \displaystyle P = \frac{1}{2} second:

d(t) = 6\cos(4\pi t)

For 3 cm below rest, solve 6\cos(4\pi t) = -3, so \displaystyle \cos(4\pi t) = -\frac{1}{2}. The first positive angle with this cosine is \displaystyle \frac{2\pi}{3}, so \displaystyle 4\pi t = \frac{2\pi}{3} and \displaystyle t = \frac{1}{6} second. At t = 0.3, d = 6\cos(1.2\pi) \approx 6(-0.8090) \approx -4.85 cm. The low turning point was at \displaystyle t = \frac{P}{2} = 0.25, so the weight is on its way back up.

Example 4

Fastest in the middle

For Jeruto’s spring, \displaystyle d(t) = -15\cos\left(\frac{5\pi}{2}t\right), find the average speed in each tenth of a second from t = 0 to t = 0.4. Then compare the average speeds over the 0.01 second just after release and the 0.01 second centered on t = 0.2.

Solution

The positions at t = 0, 0.1, 0.2, 0.3, 0.4 are

-15,\ -15\cos\tfrac{\pi}{4},\ 0,\ -15\cos\tfrac{3\pi}{4},\ 15

that is about -15, -10.61, 0, 10.61, 15. The distances covered in the four tenths are about 4.39, 10.61, 10.61 and 4.39 cm, so the average speeds are about 43.9, 106.1, 106.1 and 43.9 cm per second. The middle of the trip, near rest position, is covered more than twice as fast.

Over the first 0.01 second the weight moves \displaystyle 15 - 15\cos\frac{\pi}{40} \approx 0.046 cm, an average of about 4.6 cm per second: it has barely started. From t = 0.195 to t = 0.205 it moves from -15\cos(0.4875\pi) \approx -0.589 to about 0.589, so 1.178 cm, an average of about 117.8 cm per second. That nearly matches the fastest speed in the Results, \displaystyle |a|\omega = 15 \cdot \frac{5\pi}{2} = 37.5\pi \approx 117.8 cm per second, reached at rest position.

Example 5

A third of the time for half the trip

How long, in each cycle, is Jeruto’s weight within 7.5 cm of its rest position?

Solution

Write \displaystyle \theta = \frac{5\pi}{2}t. The weight is within 7.5 cm when |-15\cos\theta| < 7.5, that is \displaystyle |\cos\theta| < \frac{1}{2}. In one cycle, 0 \le \theta < 2\pi, this holds for \displaystyle \frac{\pi}{3} < \theta < \frac{2\pi}{3} and for \displaystyle \frac{4\pi}{3} < \theta < \frac{5\pi}{3}, two stretches of \displaystyle \frac{\pi}{3} each. Together they are \displaystyle \frac{2\pi}{3} of the 2\pi, a third of the cycle, so a third of 0.8 seconds: \displaystyle \frac{0.8}{3} \approx 0.267 seconds.

The stretch from -7.5 to 7.5 is half of each trip’s distance, yet it takes only a third of the time: the weight lingers near its turning points, where it is slow.

Example 6

Kibet’s car after a bump

After Kibet’s car hits a bump, its body bounces on the springs. With up positive, the body starts 4 cm above its rest level at t = 0, each cycle takes 0.5 second, and the shock absorbers shrink the size of the swing so that each second it keeps 40\% of its size. Write a model, find the displacement at t = 0.25, and find the first time at which the cosine factor is 1 and the body is less than 0.5 cm above rest. (The numbers are illustrative, not measured.)

Solution

Without damping this would be 4\cos(4\pi t), since \displaystyle \omega = \frac{2\pi}{0.5} = 4\pi. The swing size starts at 4 and is multiplied by 0.4 each second, so it is 4(0.4)^t after t seconds, an exponential decay with decay factor 0.4. Putting the two together,

d(t) = 4(0.4)^t\cos(4\pi t)

At t = 0.25 the cosine is \cos\pi = -1, so d = -4(0.4)^{0.25} \approx -4(0.7953) \approx -3.18 cm, the first low swing. The cosine factor is 1 at t = 0, 0.5, 1, 1.5, \ldots, where d equals 4(0.4)^t. We need 4(0.4)^t < 0.5, so \displaystyle (0.4)^t < \frac{1}{8} and, taking logarithms (the inequality flips because \ln 0.4 < 0),

t > \frac{\ln 8}{\ln 2.5} \approx 2.27

The first such time is t = 2.5, where d = 4(0.4)^{2.5} \approx 0.40 cm; at t = 2 the value is still 4(0.16) = 0.64 cm. In OpenStax’s form the damping constant is c = -\ln 0.4 \approx 0.916.

Figure 3Kibet’s car body after a bump. The swings keep the same 0.5 second rhythm while the dashed envelope 4(0.4)^t shrinks them.

Boundaries

The model is an ideal. It says the motion goes on forever, the same size each cycle. A real spring or board loses energy, mostly to friction, and comes to rest; damping is the first correction.
It needs a \ne 0 and \omega > 0; with a = 0 nothing moves.
The angle \omega t is in radians, and a frequency in hertz pairs with t in seconds.
Measuring d from equilibrium is what makes the midline 0. Measured from somewhere else, such as the floor, the model gains a constant, as in Tides, daylight and sound.
A pendulum fits this model only for small swings. This lesson gives no formula for a pendulum’s period.
In damped motion the wave touches the envelope a\,r^t where the cosine factor is 1, but its highest points come a little earlier than those touching points, and the cycle is no longer exactly periodic.

Where this goes next

Two springs moving together, or a motion that starts neither at an extreme nor at rest, give a sine plus a cosine of the same \omega t; Combining a sine and a cosine into one wave shows that this is still one wave with its own amplitude and starting point. Solving trigonometric equations answers “when is it at this position” in general. Calculus proves the speed result directly: the rate of change of a\sin(\omega t) is a\omega\cos(\omega t), largest in size at equilibrium and zero at the turning points.

Exercises

Reading a model

For d(t) = 9\cos(6\pi t) cm, give the amplitude, period and frequency, and say how the motion starts.
For d(t) = -2.5\sin(10\pi t) cm, give the amplitude, period and frequency, and say how the motion starts.
For \displaystyle d(t) = 12\cos\left(\frac{2\pi}{3}t\right) cm, give the period and frequency.
A diving board tip moves with d(t) = 0.8\sin(50\pi t) cm. Give its period and frequency.

Building a model

A weight is pushed 10 cm above rest and let go. It makes one full cycle every 1.2 seconds. Write a model, up positive.
A diving board tip at rest is struck downward at t = 0 and swings 5 cm each way at 2.5 hertz. Write a model, up positive.
A weight pulled 3 cm below rest and let go makes 30 full cycles in 18 seconds. Write a model.
A weight is lifted 7 cm and let go, and its angular frequency is 8 radians per second. Write a model and give its period and frequency.
Show that Jeruto’s model \displaystyle -15\cos\left(\frac{5\pi}{2}t\right) equals \displaystyle 15\sin\left(\frac{5\pi}{2}(t - 0.2)\right).

Positions and times

Where is Jeruto’s weight at t = 0.05 and at t = 1, and which way is it moving at t = 1?
During the first cycle, when is the tip of Chebet’s board 3 cm above rest?
For the model of Exercise 1, when does the object first reach rest position, and when is it first at its lowest?
For the model of Exercise 5, when is the weight first 5 cm below rest?
For the model of Exercise 6, during the first cycle, when is the tip more than 4 cm below rest, and for what fraction of the cycle?

Speed

For the model of Exercise 1, what is the fastest speed, where is it reached, and when first?
For the model of Exercise 1, find the average speed over \displaystyle 0 \le t \le \frac{1}{24} and over \displaystyle \frac{1}{24} \le t \le \frac{1}{12}.
Find the average speed of Jeruto’s weight over one full cycle and compare it with its fastest speed.

Damping

Another car body moves with d(t) = 5(0.5)^t\cos(2\pi t) cm. Give the period and the displacement at t = 1, 2, 3.
A weight pulled 6 cm below rest and let go has a cycle of 0.4 second, and each second its swing keeps 70\% of its size. Write a model and find d(1).
For the model of Exercise 18, after what time is the envelope 5(0.5)^t less than 0.1 cm?

Reasoning

Jeruto pulls the weight down 30 cm instead of 15 cm. According to the model, what changes and what does not? Explain from the formulas.
Use the circle picture to explain why the object is moving slowest near its turning points.
In the damped model a\,r^t\cos(\omega t) with 0 < r < 1, the envelope never reaches 0. Is that a fact about real springs? Explain.
Explain why frequency and angular frequency differ by a factor of 2\pi, and give the unit of each.

Answers

Amplitude 9 cm, \displaystyle P = \frac{2\pi}{6\pi} = \frac{1}{3} second, f = 3 hertz. It starts at its highest point, 9 cm above rest.
Amplitude 2.5 cm, \displaystyle P = \frac{2\pi}{10\pi} = \frac{1}{5} second, f = 5 hertz. It starts at rest position moving down, since a < 0.
\displaystyle P = 2\pi \div \frac{2\pi}{3} = 3 seconds and \displaystyle f = \frac{1}{3} hertz.
\displaystyle P = \frac{2\pi}{50\pi} = \frac{1}{25} = 0.04 second and f = 25 hertz.
Cosine form with a = 10 and \displaystyle \omega = \frac{2\pi}{1.2} = \frac{5\pi}{3}: \displaystyle d(t) = 10\cos\left(\frac{5\pi}{3}t\right).
Sine form, starting down, so a = -5; \omega = 2\pi \cdot 2.5 = 5\pi: d(t) = -5\sin(5\pi t).
\displaystyle P = \frac{18}{30} = 0.6 second, so \displaystyle \omega = \frac{2\pi}{0.6} = \frac{10\pi}{3} and \displaystyle d(t) = -3\cos\left(\frac{10\pi}{3}t\right).
d(t) = 7\cos(8t), \displaystyle P = \frac{2\pi}{8} = \frac{\pi}{4} \approx 0.785 second, \displaystyle f = \frac{4}{\pi} \approx 1.27 hertz.
\displaystyle \frac{5\pi}{2}(t - 0.2) = \frac{5\pi}{2}t - \frac{\pi}{2}, and \displaystyle \sin\left(\theta - \frac{\pi}{2}\right) = -\cos\theta, so \displaystyle 15\sin\left(\frac{5\pi}{2}t - \frac{\pi}{2}\right) = -15\cos\left(\frac{5\pi}{2}t\right).
\displaystyle d(0.05) = -15\cos\frac{\pi}{8} \approx -13.86 cm. \displaystyle d(1) = -15\cos\frac{5\pi}{2} = 0; t = 1 is 0.2 second into the second cycle, on the way from the bottom to the top, so it is moving up.
6\sin(3\pi t) = 3 gives \displaystyle \sin(3\pi t) = \frac{1}{2}, so \displaystyle 3\pi t = \frac{\pi}{6} or \displaystyle \frac{5\pi}{6}: \displaystyle t = \frac{1}{18} and \displaystyle t = \frac{5}{18} second.
Rest position: \displaystyle 6\pi t = \frac{\pi}{2}, \displaystyle t = \frac{1}{12} second. Lowest: 6\pi t = \pi, \displaystyle t = \frac{1}{6} second.
\displaystyle \cos\left(\frac{5\pi}{3}t\right) = -\frac{1}{2}, first at \displaystyle \frac{5\pi}{3}t = \frac{2\pi}{3}, so \displaystyle t = \frac{2}{5} = 0.4 second.
-5\sin(5\pi t) < -4 means \sin(5\pi t) > 0.8, so \arcsin 0.8 < 5\pi t < \pi - \arcsin 0.8: 0.0590 < t < 0.1410 second, about 0.0819 second of the 0.4 second cycle, about 20.5\%.
|a|\omega = 9 \cdot 6\pi = 54\pi \approx 169.6 cm per second, at rest position, first at \displaystyle t = \frac{1}{12} second.
d(0) = 9, \displaystyle d\left(\frac{1}{24}\right) = 9\cos\frac{\pi}{4} \approx 6.364, \displaystyle d\left(\frac{1}{12}\right) = 0. Distances 2.636 and 6.364 cm in \displaystyle \frac{1}{24} second each: about 63.3 and 152.7 cm per second.
One cycle covers 4 \cdot 15 = 60 cm in 0.8 second, an average of 75 cm per second, against a fastest speed of 37.5\pi \approx 117.8 cm per second.
\displaystyle P = \frac{2\pi}{2\pi} = 1 second. At t = 1, 2, 3 the cosine is 1, so d = 2.5, 1.25 and 0.625 cm.
a = -6, \displaystyle \omega = \frac{2\pi}{0.4} = 5\pi, r = 0.7: d(t) = -6(0.7)^t\cos(5\pi t). Then d(1) = -6(0.7)\cos 5\pi = 4.2 cm, above rest.
5(0.5)^t < 0.1 means 2^t > 50, so \displaystyle t > \frac{\ln 50}{\ln 2} \approx 5.64 seconds.
The amplitude doubles to 30 cm and a = -30; the fastest speed |a|\omega doubles. The period \displaystyle \frac{2\pi}{\omega} and the frequency contain no a, so they stay 0.8 second and 1.25 hertz.
The object moves like the height of a point Q going around a circle at a steady speed. Near the top or bottom of the circle, Q’s path is almost level, so its height hardly changes even though Q keeps its full speed. Those are the turning points, where the object is slowest.
No. A real spring comes to rest. The model describes the swings well while they are large enough to measure; its claim that they never end goes beyond what it was built for.
One cycle is 2\pi radians of input to the cosine. If there are f cycles each second, the input grows by 2\pi f radians each second, so \omega = 2\pi f. The unit of f is cycles per second (hertz), and the unit of \omega is radians per second.