Simple Harmonic Motion

Lessons

Senior 5 · Physics

Simple Harmonic Motion

4 lessons

Lesson 1 of 4

Recognising Simple Harmonic Motion

25 min

A clock that counts vibrations

The clock in your phone, and in almost every quartz watch, keeps time by counting the vibrations of a tiny tuning fork cut from quartz. The fork is made so that it vibrates exactly $32\,768$ times a second, which is $2^{15}$ : a chip halves that frequency fifteen times and gets one pulse per second. Why does a vibrating fork make such a good clock? Because its period does not change when its swing gets bigger or smaller. The same is true, to a good approximation, of a mass bouncing on a spring, a child on a swing, a bridge deck swaying in the wind and an atom vibrating inside a crystal. All of them move in the same mathematical way, called simple harmonic motion. In this lesson you find the single condition that makes a motion simple harmonic, and the equation that every such motion obeys.

The words for oscillation

An oscillation is a repeated to-and-fro motion about a fixed point. The fixed point is the equilibrium position: the place where the resultant force on the body is zero, and where it would stay at rest if it were not disturbed.

Quantity	Symbol and unit	Meaning
displacement	$x$ , m	the distance from the equilibrium position, with a sign that gives the direction
amplitude	$A$ , m	the greatest displacement from equilibrium (always positive)
period	$T$ , s	the time for one complete oscillation
frequency	$f$ , Hz	the number of oscillations per second, $f = \frac{1}{T}$
angular frequency	$\omega$ , $\mathrm{rad\,s^{-1}}$	$\omega = 2\pi f = \frac{2\pi}{T}$
phase	$\omega t + \phi$ , rad	the angle that tells you where in its cycle the body is

Observation: what does the force depend on?

Fix a laboratory trolley between two stretched springs, as in the diagram, and pull it sideways. When you let go it oscillates about the equilibrium position. Whichever side it is on, the springs push or pull it back towards the middle: the force is a restoring force. If you measure the force with a newton meter at several displacements, you find that doubling the displacement doubles the force. So

F = - k x

where $k$ is a positive constant (the combined stiffness of the springs, in $\mathrm{N\,m^{-1}}$ ) and the minus sign says the force always points opposite to the displacement.

Newton's second law, $F = m a$ , then gives the acceleration:

ma = -kx \quad\Longrightarrow\quad a = -\frac{k}{m}\,x

Because $\frac{k}{m}$ is positive, we can write it as the square of a positive constant, $\omega^2 = \frac{k}{m}$ . This gives the defining equation of simple harmonic motion:

a = -\omega^2 x \qquad\text{or}\qquad \frac{d^2x}{dt^2} = -\omega^2 x

A graph of $a$ against $x$ is therefore a straight line through the origin with gradient $-\omega^2$ . Any system whose acceleration graph looks like this, whatever it is made of, is a simple harmonic oscillator.

Definition

Simple harmonic motion (SHM) is the motion of a body whose acceleration is directly proportional to its displacement from a fixed point and is always directed towards that point: $a = -\omega^2 x$ , where $\omega$ is a constant called the angular frequency. Example: a trolley between two springs with $k = 50\,\mathrm{N\,m^{-1}}$ and $m = 2.0\,\mathrm{kg}$ has $a = - 25 x$ , so $\omega^2 = 25\,\mathrm{s^{-2}}$ .

Characteristics of SHM

The force, and so the acceleration, is always directed towards the equilibrium position, and is proportional to the displacement.
The acceleration is greatest ( $\omega^2 A$ ) at the two extremes, where the body is momentarily at rest, and zero at the equilibrium position, where the body moves fastest.
The motion is periodic, and its displacement–time graph is a sine or cosine curve.
The period depends only on $\omega$ , that is on the system (here $k$ and $m$ ), and not on the amplitude. Motion with this property is called isochronous, and it is why the quartz fork and the pendulum make good clocks.
Kinetic and potential energy are continually exchanged, and without friction their sum stays constant (Lesson 2).

Examples, and motions that are not SHM

System	Restoring force	SHM when
mass on a spring	spring tension or compression	the spring obeys Hooke's law
simple pendulum	component of the weight along the arc	the angle of swing is small (below about $10^\circ$ )
liquid in a U-tube	weight of the unbalanced liquid column	friction with the walls is small
floating test tube pushed down	extra upthrust	the tube has straight sides
prong of a tuning fork, quartz crystal	elastic forces in the solid	the vibration is small
atom in a crystal lattice	forces from neighbouring atoms	the vibration is small compared with the atomic spacing

Not every repeated motion is SHM. A ball bouncing on a floor is periodic, but between bounces its acceleration is $g$ downwards whatever its height, so $a$ is not proportional to $x$ . A ball rolling to and fro in a V-shaped trough feels a force of constant size, and its period grows with amplitude. Both are oscillations; neither is simple harmonic.

SHM as the shadow of circular motion

Solving $a = -\omega^2 x$ is easier if you look at something you already know. A point P moves round a circle of radius $A$ at a steady angular velocity $\omega$ . At time $t$ the radius OP makes the angle $\theta = \omega t + \phi$ with the x-axis. Drop a perpendicular from P to the diameter; its foot N moves to and fro along the diameter.

The position of N is

x = A\cos(\omega t + \phi)

The acceleration of P is the centripetal acceleration $\omega^2 A$ , directed from P towards O. The acceleration of N is the component of this along the diameter: $-\omega^2 A\cos(\omega t + \phi) = -\omega^2 x$ . The minus sign appears because the component points towards O. So N moves with SHM, its angular frequency equals the angular velocity of P, and its amplitude is the radius of the circle. The circle is called the reference circle, and this is why $\omega$ is called the angular frequency even though nothing in a spring system turns.

Checking the solution by differentiation

Differentiate $x = A\cos(\omega t + \phi)$ twice with respect to time:

\frac{dx}{dt} = -\omega A\sin(\omega t + \phi), \qquad \frac{d^2x}{dt^2} = -\omega^2 A\cos(\omega t + \phi) = -\omega^2 x

The equation is satisfied for any values of $A$ and $\phi$ . Those two constants are fixed by how the motion starts: $A$ by how far you pull the body, and $\phi$ by the instant you choose as $t = 0$ . If you start the clock as you release the body from $x = + A$ , then $\phi = 0$ and $x = A\cos\omega t$ . If you start it as the body passes through equilibrium moving in the positive direction, then $x = A\sin\omega t$ . Both describe the same motion; only the starting instant differs.

The time for one cycle is the time for P to go once round the circle:

T = \frac{2\pi}{\omega}, \qquad f = \frac{\omega}{2\pi}

Worked example 1: reading $\omega$ from the acceleration

A trolley of mass 2.0 kg is held between two springs of combined stiffness $50\,\mathrm{N\,m^{-1}}$ . Show that it moves with SHM and find its period and frequency.

Solution

The resultant force is $F = - 50 x$ , so $a = \frac{F}{m} = -\frac{50}{2.0}x = -25x$ . This has the form $a = -\omega^2 x$ with a positive constant, so the motion is simple harmonic, with $\omega^2 = 25\,\mathrm{s^{-2}}$ and $\omega = 5.0\,\mathrm{rad\,s^{-1}}$ .

$T = \frac{2\pi}{\omega} = \frac{2\pi}{5.0} = 1.26\,\mathrm{s}$ and $f = \frac{1}{T} = 0.796\,\mathrm{Hz}$ . Notice that the amplitude was not needed: the period does not depend on it.

Worked example 2: using the displacement equation

A body moves according to $x = 0.040\cos\left(4\pi t + \frac{\pi}{3}\right)$ , with $x$ in metres and $t$ in seconds. Find (a) the amplitude, frequency and period, (b) the displacement at $t = 0$ and at $t = 0.10\,\mathrm{s}$ , (c) the first time at which $x = -0.020\,\mathrm{m}$ .

Solution

(a) $A = 0.040\,\mathrm{m}$ . $\omega = 4\pi\,\mathrm{rad\,s^{-1}}$ , so $f = \frac{4\pi}{2\pi} = 2.0\,\mathrm{Hz}$ and $T = 0.50\,\mathrm{s}$ .

(b) At $t = 0$ : $x = 0.040\cos\frac{\pi}{3} = 0.040 \times 0.500 = 0.020\,\mathrm{m}$ . At $t = 0.10\,\mathrm{s}$ the phase is $0.4\pi + \frac{\pi}{3} = 2.30\,\mathrm{rad}$ (132°), so $x = 0.040\cos(2.30) = 0.040 \times (-0.669) = -0.027\,\mathrm{m}$ . Your calculator must be in radians.

(c) $\cos\left(4\pi t + \frac{\pi}{3}\right) = -0.500$ , whose smallest solution above $\frac{\pi}{3}$ is $\frac{2\pi}{3}$ . So $4\pi t = \frac{2\pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}$ , giving $t = \frac{1}{12}\,\mathrm{s} = 0.083\,\mathrm{s}$ .

Worked example 3: the quartz fork

A watch crystal vibrates at $32\,768\,\mathrm{Hz}$ . Take the amplitude at the tip of a prong as $1.0\,\mathrm{\mu m}$ (an illustrative value). Find the greatest acceleration of the tip and compare it with $g$ .

Solution

$\omega = 2\pi f = 2\pi \times 32\,768 = 2.06 \times 10^{5}\,\mathrm{rad\,s^{-1}}$ . The acceleration is greatest at the extremes: $a_{\max} = \omega^2 A = (2.06 \times 10^{5})^2 \times 1.0 \times 10^{-6} = 4.2 \times 10^{4}\,\mathrm{m\,s^{-2}}$ , about 4300 times $g$ . A tiny amplitude at a high frequency still means a huge acceleration, because $a_{\max}$ grows as $f^{2}$ .

Warning

Common mistakes. (1) Treating every periodic motion as SHM: check that $a \propto -x$ first. (2) Confusing $\omega$ with $f$ : $\omega = 2\pi f$ , so a 2.0 Hz oscillation has $\omega = 12.6\,\mathrm{rad\,s^{-1}}$ , not 2.0. (3) Using degrees in $\cos(\omega t + \phi)$ : the phase is in radians. (4) Dropping the minus sign in $a = -\omega^2 x$ ; it is what makes the force restoring.

Tip

To test for SHM, write the resultant force as a function of displacement. If it comes out as $F = -(\text{positive constant}) \times x$ , you have SHM, and $\omega^2 = \frac{\text{constant}}{m}$ . Read $\omega$ , then $T = \frac{2\pi}{\omega}$ , straight off.

Comparing Simple Harmonic Motion to Circular Motion: Demonstration (Flipping Physics). Watch a ball on a turntable and a mass on a spring side by side: their shadows move together, which is the reference circle in action.

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Practice

State the two conditions on the acceleration of a body for its motion to be simple harmonic.
A body performs SHM with period 0.40 s. Find its frequency and angular frequency.
A 0.50 kg mass on a spring has acceleration $a = - 64 x$ (SI units). Find $\omega$ , the period and the spring constant.
A body moves with $x = 0.12\sin(5.0t)$ m. Find its amplitude and period, and its displacement at $t = 0.20\,\mathrm{s}$ .
Explain why a ball bouncing vertically on a hard floor is periodic but not simple harmonic.
A prong of a 440 Hz tuning fork vibrates with amplitude 0.50 mm. Find its greatest acceleration.
A point moves round a circle of radius 0.30 m at 2.0 revolutions per second. Its shadow on a diameter performs SHM. Write an equation for the shadow's displacement, taking $x = +0.30\,\mathrm{m}$ at $t = 0$ , and find the shadow's greatest acceleration.
Show by differentiation that $x = A\sin(\omega t + \phi)$ also satisfies $\frac{d^2x}{dt^2} = -\omega^2 x$ .

Show answersHide answers

The acceleration must be directly proportional to the displacement from a fixed point, and always directed towards that point.
$f = \frac{1}{0.40} = 2.5\,\mathrm{Hz}$ ; $\omega = 2\pi \times 2.5 = 15.7\,\mathrm{rad\,s^{-1}}$ .
$\omega = \sqrt{64} = 8.0\,\mathrm{rad\,s^{-1}}$ ; $T = \frac{2\pi}{8.0} = 0.785\,\mathrm{s}$ ; $k = m\omega^2 = 0.50 \times 64 = 32\,\mathrm{N\,m^{-1}}$ .
$A = 0.12\,\mathrm{m}$ ; $T = \frac{2\pi}{5.0} = 1.26\,\mathrm{s}$ ; $x = 0.12\sin(1.0) = 0.12 \times 0.841 = 0.101\,\mathrm{m}$ .
Between bounces the only force is its weight, so its acceleration is $g$ downwards at every height: it does not depend on the displacement and does not point towards a fixed point.
$\omega = 2\pi \times 440 = 2765\,\mathrm{rad\,s^{-1}}$ ; $a_{\max} = \omega^2 A = 2765^2 \times 5.0 \times 10^{-4} = 3.8 \times 10^{3}\,\mathrm{m\,s^{-2}}$ .
$\omega = 2\pi \times 2.0 = 4\pi\,\mathrm{rad\,s^{-1}}$ , so $x = 0.30\cos(4\pi t)$ m; $a_{\max} = (4\pi)^2 \times 0.30 = 47\,\mathrm{m\,s^{-2}}$ .
$\frac{dx}{dt} = \omega A\cos(\omega t + \phi)$ ; $\frac{d^2x}{dt^2} = -\omega^2 A\sin(\omega t + \phi) = -\omega^2 x$ .

Key points

SHM: $a = -\omega^2 x$ . The restoring force is proportional to the displacement and points towards equilibrium.
Solution: $x = A\cos(\omega t + \phi)$ , with $T = \frac{2\pi}{\omega}$ independent of the amplitude.
SHM is the projection of uniform circular motion on a diameter (the reference circle).
For a mass on a spring, $\omega^2 = \frac{k}{m}$ .

Sources and credits

15.1 Simple Harmonic Motion and 15.3 Comparing Simple Harmonic Motion and Circular Motion, from University Physics Volume 1 by William Moebs, Samuel J. Ling and Jeff Sanny, OpenStax, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: 32768 Hz quartz crystal resonator, Mister rf, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: Comparing Simple Harmonic Motion (SHM) to Circular Motion - Demonstration, Flipping Physics, embedded from YouTube.

This lesson adapts CC BY-SA material and is shared under CC BY-SA 4.0.

The four diagrams, the worked examples and the practice exercises are written by Ganzaa.

1A clock that counts vibrations

2The words for oscillation

3Observation: what does the force depend on?

4Characteristics of SHM

Examples, and motions that are not SHM

5SHM as the shadow of circular motion

Checking the solution by differentiation

A clock that counts vibrations

The words for oscillation

Observation: what does the force depend on?

Characteristics of SHM

SHM as the shadow of circular motion