Unit 4, lesson 57

Harmonic Motion

How a sine or cosine function describes a weight bouncing on a spring, what amplitude, period and phase mean for that motion, and how an exponential factor models swings that fade.

  • Theory part 113
  • Practice part 114
  • Oscillation

Hang a weight on a spring, pull it down a little and let go. It bobs up and down through the spot where it hung at rest, and every bounce takes the same time. Plot its height against time and you get a sine wave. The same picture describes the height of a seat on a turning Ferris wheel and, to a good approximation, a pendulum swinging through a small angle.

Key terms

Equilibrium position

Definition The equilibrium position of an object that can move back and forth along a line is a position where it can stay at rest, because the forces on it balance. For a weight hanging on a spring it is the point where the upward pull of the stretched spring exactly balances the downward pull of gravity. In a model of the motion, positions are measured from this point, so the equilibrium position is displacement 00.

In plain words It is where the object would sit if nothing disturbed it. Every oscillation below swings around such a resting point.

Example A spring with spring constant k=50k = 50 newtons per meter pulls back with 5050 newtons for each meter it is stretched. Hang a weight of mass 0.50.5 kilograms on it. Gravity pulls down with 0.5⋅9.8=4.90.5 \cdot 9.8 = 4.9 newtons, so the weight comes to rest when the spring pulls up with 4.94.9 newtons. That happens at a stretch of

4.950=0.098 meters.\frac{4.9}{50} = 0.098 \text{ meters.}

So the equilibrium position is 9.89.8 centimeters below the end of the unloaded spring.

Displacement

Definition Choose a positive direction along the line of motion. The displacement of the object at time tt, written d(t)d(t), is its signed distance from the equilibrium position, positive on the chosen side and negative on the other. If positions are measured as y(t)y(t) on a coordinate line that points in the chosen positive direction, and the equilibrium position is y=Dy = D, then

d(t)=y(t)−D.d(t) = y(t) - D.

The distance from equilibrium is ∣d(t)∣\lvert d(t) \rvert.

In plain words Displacement tells you how far the object is from its resting point and on which side. The sign gives the side and the size gives the distance.

Example Measure heights above the floor with up as the positive direction, and suppose the weight rests 5050 centimeters above the floor. When it is 5656 centimeters up, its displacement is 56−50=656 - 50 = 6 centimeters. When it is 4747 centimeters up, its displacement is 47−50=−347 - 50 = -3 centimeters, so it is 33 centimeters below the resting point.

Simple harmonic motion

Definition An object moves in simple harmonic motion if its displacement from equilibrium at time tt is

d(t)=asin⁡(ωt+φ),d(t) = a\sin(\omega t + \varphi),

where aa, ω\omega and φ\varphi are real constants with a≠0a \ne 0 and ω>0\omega > 0, and the angle ωt+φ\omega t + \varphi is in radians. The most common cases are d(t)=asin⁡(ωt)d(t) = a\sin(\omega t) and d(t)=acos⁡(ωt)d(t) = a\cos(\omega t), which is the case φ=π2\varphi = \frac{\pi}{2} because cos⁡x=sin⁡(x+π2)\cos x = \sin\left(x + \frac{\pi}{2}\right) for every real xx. Physics shows that an object moves this way when the force pulling it back toward equilibrium is proportional to its displacement and nothing slows it down, as for a weight on an ideal spring without friction.

In plain words The object swings through its resting point to the same distance on each side, again and again, and every full swing takes the same time.

Example Suppose the displacement is d(t)=4cos⁡(πt)d(t) = 4\cos(\pi t) centimeters after tt seconds, with up as positive.

tt000.50.5111.51.522
d(t)d(t)4400−4-40044

The object starts 44 centimeters above equilibrium, passes through equilibrium at t=0.5t = 0.5, reaches its lowest point at t=1t = 1 and is back at the top at t=2t = 2. Between these times it moves smoothly. For instance, a quarter second in, it is at 4cos⁡π4=22≈2.834\cos\frac{\pi}{4} = 2\sqrt{2} \approx 2.83 centimeters.

Amplitude

Definition The amplitude of the motion d(t)=asin⁡(ωt+φ)d(t) = a\sin(\omega t + \varphi) is ∣a∣\lvert a \rvert. For every tt

−∣a∣≤d(t)≤∣a∣,-\lvert a \rvert \le d(t) \le \lvert a \rvert,

and both bounds are reached, so the amplitude is the largest distance from equilibrium. In simple harmonic motion the amplitude has no effect on the period.

In plain words Amplitude is how far the object gets from its resting point. A bigger pull gives a bigger swing, but in this model not a slower one.

Example Pull the weight from the spring example 66 centimeters below equilibrium and release it from rest at t=0t = 0. It swings up to 66 centimeters above equilibrium and back, so the amplitude is 66 centimeters. With up as positive, a start at the lowest point gives the model

d(t)=−6cos⁡(10t) centimeters,d(t) = -6\cos(10t) \text{ centimeters,}

where the 1010 comes from the next section. The coefficient is −6-6, but the amplitude is ∣−6∣=6\lvert -6 \rvert = 6. A pull of only 33 centimeters would give d(t)=−3cos⁡(10t)d(t) = -3\cos(10t), a smaller swing with exactly the same timing.

Period and frequency

Definition The period of d(t)=asin⁡(ωt+φ)d(t) = a\sin(\omega t + \varphi) is the time TT of one full cycle, and the frequency ff is the number of cycles per unit of time,

T=2πω,f=1T=ω2π.T = \frac{2\pi}{\omega}, \qquad f = \frac{1}{T} = \frac{\omega}{2\pi}.

The period has this value because the angle ωt+φ\omega t + \varphi grows by 2π2\pi exactly when tt grows by 2πω\frac{2\pi}{\omega}. The constant ω\omega is the angular frequency, in radians per unit of time. With tt in seconds, frequency is measured in cycles per second, also called hertz. For a weight of mass mm on an ideal spring with spring constant kk, physics gives

ω=km.\omega = \sqrt{\frac{k}{m}}.

In plain words The period is how long one full bounce takes, down and back up. The frequency counts bounces per second. A stiffer spring or a lighter weight makes the bounces quicker.

Example For the spring with k=50k = 50 newtons per meter and the 0.50.5 kilogram weight,

ω=500.5=10 radians per second,T=2π10=π5≈0.63 seconds.\begin{aligned} \omega &= \sqrt{\frac{50}{0.5}} \\[0.8em] &= 10 \text{ radians per second}, \\[0.8em] T &= \frac{2\pi}{10} = \frac{\pi}{5} \approx 0.63 \text{ seconds.} \end{aligned}

The frequency is 5π≈1.59\frac{5}{\pi} \approx 1.59 cycles per second, so the weight bounces a little more than three times every two seconds.

Phase

Definition In d(t)=asin⁡(ωt+φ)d(t) = a\sin(\omega t + \varphi) the angle ωt+φ\omega t + \varphi is the phase of the motion at time tt, and the constant φ\varphi is the phase constant, the phase at t=0t = 0. Adding a multiple of 2π2\pi to φ\varphi does not change the motion, so the phase constant is determined only up to such a multiple. Writing

ωt+φ=ω(t+φω)\omega t + \varphi = \omega\left(t + \frac{\varphi}{\omega}\right)

shows that the graph of dd is the graph of asin⁡(ωt)a\sin(\omega t) shifted by ∣φ∣ω\frac{\lvert \varphi \rvert}{\omega} time units, to the left when φ>0\varphi > 0 and to the right when φ<0\varphi < 0. So the phase shift, the number CC in Asin⁡(B(t−C))+DA\sin\bigl(B(t - C)\bigr) + D, is C=−φωC = -\frac{\varphi}{\omega}, which is a different number from the phase constant φ\varphi. Two motions with the same ω\omega and with a>0a > 0 in both are in phase when their phase constants differ by a multiple of 2π2\pi, so they reach their highest points at the same moments. They are opposite in phase when their phase constants differ by π\pi plus a multiple of 2π2\pi, since sin⁡(x+π)=−sin⁡x\sin(x + \pi) = -\sin x.

In plain words The phase tells you where in its cycle the object is. The phase constant tells you where in the cycle it was when the clock started.

Example Take d(t)=3sin⁡(2t+π3)d(t) = 3\sin\left(2t + \frac{\pi}{3}\right). At t=0t = 0 the phase is π3\frac{\pi}{3}, so the object starts at

d(0)=3sin⁡π3=332≈2.60,d(0) = 3\sin\frac{\pi}{3} = \frac{3\sqrt{3}}{2} \approx 2.60,

partway up and still rising toward its maximum 33. It reaches the top when the phase is π2\frac{\pi}{2}, that is when 2t+π3=π22t + \frac{\pi}{3} = \frac{\pi}{2}, at t=π12≈0.26t = \frac{\pi}{12} \approx 0.26. Since 2t+π3=2(t+π6)2t + \frac{\pi}{3} = 2\left(t + \frac{\pi}{6}\right), the graph is the graph of 3sin⁡(2t)3\sin(2t) shifted π6\frac{\pi}{6} to the left. The phase shift is −π6-\frac{\pi}{6}, while the phase constant is π3\frac{\pi}{3}.

Sinusoidal model

Definition A sinusoidal model of a quantity that oscillates around a fixed level is a function

y(t)=Asin⁡(B(t−C))+Dory(t)=Acos⁡(B(t−C))+D\begin{gathered} y(t) = A\sin\bigl(B(t - C)\bigr) + D \\[0.4em] \text{or} \\[0.4em] y(t) = A\cos\bigl(B(t - C)\bigr) + D \end{gathered}

where AA, BB, CC and DD are real constants with A≠0A \ne 0 and B>0B > 0, and tt is time. When y(t)y(t) is the position of an object in simple harmonic motion, the midline DD is the equilibrium position, ∣A∣\lvert A \rvert is the amplitude, BB is the angular frequency ω\omega and y(t)−Dy(t) - D is the displacement. To build a model, read off the equilibrium, the amplitude and the period, as in amplitude, period, and phase shift. Then choose sine or cosine, the sign of AA and the shift CC so that the model starts in the right place.

In plain words The displacement swings around zero. The position swings around the equilibrium, so you add the equilibrium as the midline.

Example The weight on the spring rests 5050 centimeters above the floor. It is pulled down to 4444 centimeters and released at t=0t = 0, and its period is π5\frac{\pi}{5} seconds. The equilibrium gives D=50D = 50, the amplitude is 66, and B=2π÷π5=10B = 2\pi \div \frac{\pi}{5} = 10. A negative cosine starts at its lowest point, just like the weight, so its height in centimeters is

h(t)=50−6cos⁡(10t).h(t) = 50 - 6\cos(10t).
tt00π20\frac{\pi}{20}π10\frac{\pi}{10}3π20\frac{3\pi}{20}π5\frac{\pi}{5}
h(t)h(t)44445050565650504444

After one second the model gives h(1)=50−6cos⁡10≈55.0h(1) = 50 - 6\cos 10 \approx 55.0 centimeters. A real spring loses a little energy on every bounce, and the damping model in the next section accounts for that.

Damping and the exponential envelope

Definition In damped harmonic motion, friction makes the swings shrink over time. A standard model is

d(t)=ae−ctcos⁡(ωt)d(t) = a e^{-ct}\cos(\omega t)

with constants a>0a > 0, c>0c > 0 and ω>0\omega > 0. Since ∣cos⁡(ωt)∣≤1\lvert \cos(\omega t) \rvert \le 1 and e−ct>0e^{-ct} > 0,

−ae−ct≤d(t)≤ae−ctfor every t,\begin{gathered} -a e^{-ct} \le d(t) \le a e^{-ct} \\[0.4em] \text{for every } t, \end{gathered}

so the graph stays between the exponential curves y=ae−cty = a e^{-ct} and y=−ae−cty = -a e^{-ct}, called the envelope. It touches the upper curve where cos⁡(ωt)=1\cos(\omega t) = 1 and the lower curve where cos⁡(ωt)=−1\cos(\omega t) = -1, and it passes through equilibrium where cos⁡(ωt)=0\cos(\omega t) = 0, once every πω\frac{\pi}{\omega} time units. Over each time span of length 2πω\frac{2\pi}{\omega} the envelope shrinks by the factor e−2πc/ωe^{-2\pi c / \omega}. Setting c=0c = 0 gives back simple harmonic motion.

In plain words The cosine supplies the back and forth, and the exponential factor shrinks the swings the same way exponential decay shrinks a quantity. The motion keeps its rhythm while it fades.

Example Take d(t)=10e−0.2tcos⁡(2πt)d(t) = 10e^{-0.2t}\cos(2\pi t) centimeters with tt in seconds. At every whole second cos⁡(2πt)=1\cos(2\pi t) = 1, so the graph touches the upper envelope there.

tt0011223355
d(t)d(t)10108.198.196.706.705.495.493.683.68

From one whole second to the next the swing is multiplied by e−0.2≈0.82e^{-0.2} \approx 0.82. The envelope drops below 11 centimeter when 10e−0.2t<110e^{-0.2t} < 1, that is when

t>ln⁡100.2=5ln⁡10≈11.5 seconds.t > \frac{\ln 10}{0.2} = 5\ln 10 \approx 11.5 \text{ seconds.}

Check yourself

  1. A weight on a spring has displacement d(t)=8sin⁡(4πt)d(t) = 8\sin(4\pi t) centimeters after tt seconds. Find the amplitude, the period and the frequency, and find the displacement at t=18t = \frac{1}{8} and at t=13t = \frac{1}{3}.
Show the answer

The amplitude is 88 centimeters. With ω=4π\omega = 4\pi the period is 2π4π=12\frac{2\pi}{4\pi} = \frac{1}{2} second and the frequency is 22 cycles per second. The displacements are

d(18)=8sin⁡π2=8,d(13)=8sin⁡4π3=−43≈−6.93.\begin{gathered} d\left(\tfrac{1}{8}\right) = 8\sin\frac{\pi}{2} = 8, \\[0.8em] d\left(\tfrac{1}{3}\right) = 8\sin\frac{4\pi}{3} = -4\sqrt{3} \approx -6.93. \end{gathered}

At the first time the weight is at its highest point, and at the second it is below equilibrium.

  1. A Ferris wheel has radius 2020 meters, its center is 2222 meters above the ground, and it turns once every 44 minutes. A rider boards at the lowest point at t=0t = 0. Write the rider’s height h(t)h(t) in meters after tt minutes and check it at the top.
Show the answer

The midline is 2222, the amplitude is 2020 and B=2π4=π2B = \frac{2\pi}{4} = \frac{\pi}{2}. The rider starts at the bottom, so

h(t)=22−20cos⁡(πt2).h(t) = 22 - 20\cos\left(\frac{\pi t}{2}\right).

At t=0t = 0 the height is 22−20=222 - 20 = 2 meters. Half a turn later, at t=2t = 2, it is 22−20cos⁡π=4222 - 20\cos\pi = 42 meters, the top of the wheel.

  1. A damped oscillation is modeled by d(t)=6e−0.1tcos⁡(πt)d(t) = 6e^{-0.1t}\cos(\pi t) centimeters after tt seconds. When has the envelope shrunk to half its starting size, and what is the displacement at t=2t = 2?
Show the answer

The envelope 6e−0.1t6e^{-0.1t} equals 33 when e−0.1t=12e^{-0.1t} = \frac{1}{2}, so

t=ln⁡20.1=10ln⁡2≈6.93 seconds.t = \frac{\ln 2}{0.1} = 10\ln 2 \approx 6.93 \text{ seconds.}

At t=2t = 2 the cosine is cos⁡2π=1\cos 2\pi = 1, so the displacement is 6e−0.2≈4.916e^{-0.2} \approx 4.91 centimeters, right on the upper envelope.

How this lesson connects