David Bauer Physics & Astronomy · UCLA

Physics 5B Chapter 15

Simple Harmonic Motion

Periodic motion

Period and frequency

We now begin a new unit on oscillations and waves. We start with motion that repeats itself at regular time intervals, which is called periodic motion.

The time it takes for the motion to repeat is called the period $T$. Each full repetition of the motion is called a cycle. The number of cycles per unit time is the frequency $f$ of the motion. The frequency and period are related by

\[f = \frac{1}{T}.\]

The unit of frequency is the hertz: $1\ \mathrm{Hz} = 1\ \mathrm{s^{-1}}$, or one cycle per second.

Periodic motion in living things

Living things are full of periodic motion. The table gives some typical frequencies and periods.

Motion Frequency Period
Breathing at rest $0.25\ \mathrm{Hz}$ (15 breaths per minute) $4\ \mathrm{s}$
Resting heartbeat $1.2\ \mathrm{Hz}$ (72 beats per minute) $0.8\ \mathrm{s}$
Swing of a leg while walking about $1\ \mathrm{Hz}$ about $1\ \mathrm{s}$
Hummingbird wingbeat about $50\ \mathrm{Hz}$ about $20\ \mathrm{ms}$
Vibration of your vocal folds $100$ to $250\ \mathrm{Hz}$ $4$ to $10\ \mathrm{ms}$
Mosquito wingbeat about $500\ \mathrm{Hz}$ about $2\ \mathrm{ms}$

Periodic motion only has to repeat. A heartbeat repeats, but the shape of each beat is complicated. In this chapter we will study the simplest kind of periodic motion. It describes a mass on a spring, a pendulum swinging through small angles, and, as we will see in the chapter on sinusoidal waves, each point on a string carrying a sinusoidal wave.

Example: Your heartbeat

You measure your resting pulse and count 66 beats in one minute. What are the frequency and the period of your heartbeat?

Solution

The frequency is the number of beats per second:

\[f = \frac{66\ \text{beats}}{60\ \mathrm{s}} = 1.10\ \mathrm{Hz}.\]

The period is the time for one beat:

\[T = \frac{1}{f} = 0.909\ \mathrm{s}.\]

Uniform circular motion

You have already studied one kind of periodic motion in Physics 5A: uniform circular motion (UCM), motion in a circle at constant speed. In UCM, the period $T$ is the time for one revolution, and the frequency $f = 1/T$ is the number of revolutions per unit time.

In one period, the object travels one circumference, so its speed is

\[v = \frac{2\pi r}{T} = 2\pi r f.\]

Circular motion will give us a way to describe oscillations, so let’s review it, this time describing the object’s position by an angle.

Angular position

We can describe the position of an object on a circle by giving the angle $\phi$ between the radius to the object and a reference direction. We will always measure $\phi$ counterclockwise from the $+x$-axis.

We measure $\phi$ in radians. One full revolution is $\phi = 2\pi$ (equivalent to $360^\circ$), so adding $2\pi$ to $\phi$ brings the object back to the same point.

A circle of radius r centered at the origin of x and y axes. A blue dot on the circle marks an object in uniform circular motion. The radius from the center to the object makes an angle phi with the positive x-axis, measured counterclockwise. A blue arrow just outside the circle shows the object moving counterclockwise.

Angular frequency

In UCM, the angle $\phi$ increases at a steady rate. We call this rate the angular frequency $\omega$:

\[\omega = \frac{\Delta\phi}{\Delta t}.\]

In one period $\Delta t = T$, the angle increases by one revolution, $\Delta\phi = 2\pi$, so

\[\boxed{\omega = \frac{2\pi}{T},}\qquad \text{ or }\qquad \boxed{\omega = 2\pi f.}\]

To get the second equation we substitute $1/T = f$.

The frequency $f$ counts cycles per second. The angular frequency $\omega$ counts radians per second, and there are $2\pi$ radians in each cycle. We give $\omega$ in $\mathrm{rad/s}$, never in $\mathrm{Hz}$.

The angle as a function of time

Suppose the object starts at angle $\phi_0$ at $t = 0$. After a time $t$, the angle has increased by $\omega t$, so

\[\phi(t) = \omega t + \phi_0.\]

This has the same form as uniform motion, $x(t) = v_x t + x_0$, with the angle in place of the position.

A circle centered at the origin of x and y axes. A hollow dot on the circle, labeled t equals 0, sits at a small angle phi sub 0 above the positive x-axis. A filled blue dot, labeled time t, sits farther counterclockwise on the circle, in the second quadrant. A small arc near the center marks phi sub 0 from the x-axis to the starting position, and a larger arc marks the additional angle omega t swept from the starting position to the position at time t.

The mass-spring system

A mass on a spring

Our model oscillating system is a block of mass $m$ attached to an ideal spring and sliding on a frictionless horizontal surface. We take the $x$-axis along the spring, with $x = 0$ at the equilibrium position, where the spring is relaxed.

We will write the spring constant as $k_s$. When we study waves, we will need the letter $k$ for something else.

A block of mass m on a frictionless horizontal floor, attached by a stretched coil spring to a wall on its left. A dashed outline shows where the block sits at equilibrium, where the spring is relaxed; a dashed vertical line through it marks x equals 0. The block is pulled to the right of equilibrium, to a position x, as marked on an x-axis below the floor. A red force arrow labeled F sub el, the spring force, points to the left, back toward equilibrium.

Newton’s second law for the mass

The spring exerts a restoring force: it always pulls or pushes the block back toward equilibrium. By Hooke’s law, its $x$-component is $F_{\mathrm{el},x} = -k_s x$. The spring force is the net force on the block, so Newton’s second law gives

\[-k_s x = m a_x.\]

The acceleration $a_x = -(k_s/m)\,x$ depends on where the block is, so it changes as the block moves. The constant-acceleration kinematic equations from Physics 5A do not apply to this motion.

To describe the motion of a mass on a spring, we need to find a position function $x(t)$ such that the acceleration is always proportional to the negative of the position.

What does the motion look like?

Instead of solving for $x(t)$ directly, let’s look at some data. Here is the position of a mass oscillating on a spring, measured from its equilibrium position. (This spring was vertical rather than horizontal, but that doesn’t affect our answer.)

A scatter plot of measured displacement x in centimeters, from minus 1.5 to 1.5, versus time t in seconds, from 0 to 2.7. The data points trace a smooth, regular wave. The displacement starts at a maximum of about 1.2 centimeters at t equals 0, falls through zero to a minimum of about minus 1.3 centimeters, and rises back to a maximum at about 0.8 seconds. The pattern repeats a little more than three times, with the same shape each cycle and crests that vary slightly in height, between about 1.1 and 1.35 centimeters, and troughs between about minus 1.2 and minus 1.4 centimeters.

An educated guess

The data look like a cosine curve. So let’s guess that the motion has the form

\[x = A\cos\phi.\]

(Physicists use the German word ansatz for such a guess, probably because that sounds more clever than “guess.”)

The amplitude $A$ is the maximum displacement from equilibrium. The position swings between $x = +A$ and $x = -A$.

The angle $\phi$ is called the phase. It increases with time, and each time it increases by $2\pi$, the motion completes one cycle.

The same measured displacement versus time data, with a smooth cosine curve of amplitude 1.26 centimeters and period 0.806 seconds drawn through the points; the curve follows the data closely. A vertical dimension arrow labeled A runs from x equals 0 up to the height of a crest, marking the amplitude. A horizontal dimension arrow labeled T spans from the first crest to the second crest, at about 0.8 seconds, marking one period.

Simple harmonic motion and circular motion

We can interpret $A\cos\phi$ as the $x$-coordinate of a point at angle $\phi$ on a circle of radius $A$ that we call the reference circle. As the point goes around the circle at a steady rate, its projection onto the $x$-axis moves back and forth exactly like the block.

Top: a reference circle of radius A centered on a horizontal x axis. A blue state point on the circle sits at angle phi above the positive x direction, and the radius to it is labeled A. A red line segment along the x-axis, from the center to directly below the state point, is labeled A cos phi: it is the projection of the point onto the x-axis. Bottom: a block on a spring, attached to a wall at the left, sits directly below the projection. A dashed vertical line connects the state point, its projection, and the center of the block, showing that the block's position x equals A cos phi. An x axis below the floor has tick marks at minus A, 0, and plus A.

The phase changes with time

There is no physical circular motion here. The phase $\phi$ is the angle of a point on the reference circle, which is a mathematical picture of the motion rather than a physical circle.

Just as in uniform circular motion, the phase increases at a steady angular frequency $\omega$:

\[\phi = \omega t + \phi_0.\]

The phase constant $\phi_0$ is the phase at $t = 0$. Substituting this into $A\cos\phi$ gives the position of the block at any time:

\[\boxed{x(t) = A\cos(\omega t + \phi_0).}\]

We call this motion that oscillates sinusoidally simple harmonic motion (SHM).

Derivation: Velocity and acceleration

We get the velocity and acceleration by taking derivatives with respect to time. By the chain rule, the derivative of $\cos(\omega t + \phi_0)$ is $-\omega\sin(\omega t + \phi_0)$, so the velocity is

\[v_x(t) = \frac{dx}{dt} = -A\omega\sin(\omega t + \phi_0).\]

The derivative of $\sin(\omega t + \phi_0)$ is $\omega\cos(\omega t + \phi_0)$, so differentiating once more gives the acceleration:

\[a_x(t) = \frac{dv_x}{dt} = -A\omega^2\cos(\omega t + \phi_0).\]

The defining feature of simple harmonic motion

If we compare the acceleration with the displacement $x(t) = A\cos(\omega t + \phi_0)$, we see that

\[a_x(t) = -\omega^2 \left[A\cos(\omega t + \phi_0)\right] = -\omega^2 x(t).\]

That is, the acceleration is proportional to minus the displacement:

\[\boxed{a_x = -\omega^2 x.}\]

This is the defining feature of SHM. Whenever an object’s acceleration is a negative constant times its displacement, we know it moves in SHM, and the proportionality constant is $\omega^2$.

The angular frequency of a mass on a spring

From Newton’s second law, we have $-k_s x = m a_x$. If we plug in $a_x = -\omega^2 x$, we get

\[-k_s x = -m\omega^2 x.\]

For our guessed solution to obey Newton’s second law, we need $\omega^2 = k_s/m$, so the angular frequency must be

\[\boxed{\omega = \sqrt{\frac{k_s}{m}}.}\]

For a given mass, a stiffer spring (larger $k_s$) gives a higher oscillation rate. For a given spring constant, a greater mass (larger $m$) gives a smaller rate.

Note that $\omega$ does not depend on $A$, so the oscillation rate is independent of the amplitude. This is always true for SHM.

Position, velocity, and acceleration

Here are $x$, $v_x$, and $a_x$ over two cycles for $\phi_0 = 0$. Dashed lines mark the turning points, where $x = \pm A$.

Three stacked graphs sharing a time axis that runs over two periods, from 0 to 2T, with tick marks at T over 2, T, 3T over 2, and 2T. Top graph: position x versus t, a blue cosine curve starting at its maximum plus A at t equals 0, reaching minus A at T over 2, and returning to plus A at T. Middle graph: velocity v sub x versus t, a green curve starting at zero, dipping to minus A omega at T over 4, and back to zero at T over 2; it is an upside-down sine curve. Bottom graph: acceleration a sub x versus t, an orange curve that is the mirror image of the position curve, starting at minus A omega squared, where the position is plus A. The vertical axis at t equals 0 and dashed vertical lines at T over 2, T, 3T over 2, and 2T mark the turning points: at each one, the velocity is zero and the acceleration has its largest magnitude, opposite in sign to the position.

Maximum speed and acceleration

The sine and cosine functions range between $-1$ and $+1$. So the maximum speed and the maximum magnitude of the acceleration in SHM are

\[v_{\mathrm{max}} = A\omega, \qquad a_{\mathrm{max}} = A\omega^2.\]

The block moves fastest as it passes through equilibrium, where $x = 0$ and the spring exerts no force. At the turning points, $x = \pm A$, the block is momentarily at rest, and the spring force and acceleration are largest.

Summary: the mass-spring oscillator

A mass $m$ on a spring with spring constant $k_s$ moves in simple harmonic motion with angular frequency $\omega = \sqrt{k_s/m}$. Its position, velocity, and acceleration are

\[\begin{aligned} x(t) & = A\cos(\omega t + \phi_0), \\ v_x(t) & = -A\omega\sin(\omega t + \phi_0), \\ a_x(t) & = -A\omega^2\cos(\omega t + \phi_0). \end{aligned}\]

The maximum speed is $v_{\mathrm{max}} = A\omega$, reached at equilibrium, and the maximum magnitude of the acceleration is $a_{\mathrm{max}} = A\omega^2$, reached at the turning points.

Example: A block on a spring

A $0.500\ \mathrm{kg}$ block on a frictionless horizontal surface is attached to a spring with spring constant $20.0\ \mathrm{N/m}$. You pull the block $10.0\ \mathrm{cm}$ to the right of equilibrium and release it from rest at $t = 0$.

  1. What are the angular frequency and the period of the motion?
  2. What is the block’s maximum speed?
  3. What is the maximum magnitude of the block’s acceleration?
  4. Write an expression for the block’s position $x(t)$, taking $+x$ to the right.
Solution

1. The angular frequency is set by the spring and the mass:

\[\omega = \sqrt{\frac{k_s}{m}} = \sqrt{\frac{20.0\ \mathrm{N/m}}{0.500\ \mathrm{kg}}} = 6.32\ \mathrm{rad/s}, \qquad T = \frac{2\pi}{\omega} = 0.993\ \mathrm{s}.\]

2. The block is released from rest $10.0\ \mathrm{cm}$ from equilibrium, so that is a turning point and the amplitude is $A = 0.100\ \mathrm{m}$. The maximum speed is

\[v_{\mathrm{max}} = A\omega = (0.100\ \mathrm{m})(6.32\ \mathrm{rad/s}) = 0.632\ \mathrm{m/s}.\]

3. The maximum magnitude of the acceleration is

\[a_{\mathrm{max}} = A\omega^2 = \frac{A k_s}{m} = \frac{(0.100\ \mathrm{m})(20.0\ \mathrm{N/m})}{0.500\ \mathrm{kg}} = 4.00\ \mathrm{m/s^2}.\]

4. At $t = 0$, the block is at $x = +A$. Since $x(0) = A\cos\phi_0$, we need $\cos\phi_0 = 1$, so $\phi_0 = 0$:

\[x(t) = (10.0\ \mathrm{cm})\cos\!\big[(6.32\ \mathrm{rad/s})\,t\big].\]

With our cosine form, releasing an oscillator from rest at its positive turning point always gives $\phi_0 = 0$. (The radian in $\mathrm{rad/s}$ is a reminder that $\omega t$ is an angle. Multiplying by a time in seconds cancels the seconds and leaves an angle in radians.)

Phase in simple harmonic motion

The phase determines the state

The phase $\phi = \omega t + \phi_0$ tells where the oscillator is in its cycle. For a given oscillator, once you know $\phi$, you know both the position and the velocity:

\[x = A\cos\phi, \qquad v_x = -A\omega\sin\phi.\]

Knowing only the position is not enough. Between the turning points, the block passes through each position twice in every cycle, once moving left and once moving right. The phase is different at those two instants.

A reference circle of radius A centered on a horizontal x axis. Two state points sit on the circle, one at phase pi over 3 in the upper half and one at phase 5 pi over 3 in the lower half, directly above and below each other. Both project to the same position on the x-axis, marked by a hollow dot. A green velocity arrow just above the axis at that position points to the left: at phase pi over 3 the block moves in the minus x direction. A green velocity arrow just below the axis points to the right: at phase 5 pi over 3 the block moves in the plus x direction. Each phase label sits beside its state point.

Special values of the initial phase

The initial phase $\phi_0$ tells us where in its cycle the oscillator starts at $t = 0$. Here are three important special cases:

  • $\phi_0 = 0$: the oscillator starts at $x = +A$, at rest.
  • $\phi_0 = \pi/2$: the oscillator starts at $x = 0$, moving in the $-x$ direction.
  • $\phi_0 = \pi$: the oscillator starts at $x = -A$, at rest.

Three panels side by side, each with a small reference circle above a graph of x versus t. Left panel, phi sub 0 equals 0: the state point starts on the positive x-axis, and the x versus t graph is a cosine that starts at its maximum plus A. The oscillator starts at x equals plus A, at rest. Middle panel, phi sub 0 equals pi over 2: the state point starts at the top of the circle, and the graph starts at zero and decreases, like an upside-down sine curve. The oscillator starts at x equals 0, moving in the minus x direction at maximum speed. Right panel, phi sub 0 equals pi: the state point starts on the negative x-axis, and the graph starts at its minimum minus A. The oscillator starts at x equals minus A, at rest.

Check your understanding:

  1. At $t = 0$, a block on a spring passes through equilibrium moving in the $+x$ direction. Where on the reference circle does the state point start, and what is $\phi_0$?
  2. Two identical mass-spring systems oscillate with the same amplitude. One has $\phi_0 = 0$ and the other has $\phi_0 = \pi$. Describe how the motions of the two blocks compare.
Answer

1. The state point must start at the bottom of the reference circle, so $\phi_0 = 3\pi/2$ (equivalently, $-\pi/2$). Then $x(0) = A\cos(3\pi/2) = 0$ and $v_x(0) = -A\omega\sin(3\pi/2) = +A\omega$.

2. Their phases always differ by $\pi$, so $x_2(t) = -x_1(t)$. The blocks are always on opposite sides of equilibrium, at equal distances, moving in opposite directions.

Phase space

The reference circle is not a physical circle. The horizontal coordinate of the state point is the position, $x = A\cos\phi$. Its vertical coordinate is $A\sin\phi = -v_x/\omega$, which encodes the velocity.

Phase space for a mass on a spring. The horizontal axis is position x and the vertical axis is minus v sub x over omega. A circle of radius A centered at the origin is traced counterclockwise by a blue state point, here at phase phi in the first quadrant. Dashed guide lines drop from the state point to the horizontal axis, at x equals A cos phi, and across to the vertical axis, at A sin phi.

On the upper half of the circle, the block moves in the $-x$ direction. On the lower half, it moves in the $+x$ direction.

Each point on the circle stands for one state of the oscillator (its position together with its velocity). The space of all possible states is called phase space.

We saw this same idea in the context of $pV$ diagrams. A point on a $pV$ diagram is not a physical location. It represents the state of a gas, and a process is a path from one state to another. A point in phase space likewise represents the state of the oscillator, and its motion over time is a path in phase space. For SHM, that path is a circle.

In both pictures, the size of a curve tells you something physical. For a given amount of gas, an isotherm at higher temperature lies farther from the origin of a $pV$ diagram. For a given mass-spring system, a larger amplitude traces a larger circle in phase space, and we will see in the next chapter that it has more energy.