C.1Simple harmonic motion + HL extra

Oscillations are everywhere: a swing, a guitar string, a heartbeat, the atoms in a solid, the quartz crystal in a watch. One special kind, simple harmonic motion (SHM), is both very common and mathematically simple. It is also the building block for all of wave physics. Sections 1–5 are for everyone. Sections 6–7 are HL only and clearly marked.

Knowledge and science

Nature of science

ModelsObservationsMeasurementScience as a shared endeavour

Models. Real oscillations are messy, but SHM is a model that captures their essentials. Even complicated repeating motions can be built by adding together simple sine-shaped oscillations, an idea developed by Joseph Fourier in the early 1800s.

Observations. Galileo is said to have noticed that a swinging cathedral lamp took the same time for each swing, however far it swung. He timed it against his own pulse. Oscillations whose period doesn't depend on amplitude are called isochronous. They made accurate pendulum clocks possible.

Measurement and a shared endeavour. Time itself is now defined by an oscillation: the second is a fixed number of cycles of radiation from caesium atoms. Countries agree on this standard, which keeps GPS, power grids and the internet in sync.

ToK: questions to think about

When is a simple model good enough? A pendulum is only simple harmonic for small swings, but we use the SHM model anyway. How do scientists decide when a simple model is no longer detailed enough?
What counts as proof? Fourier's work was criticised because parts of it relied on intuition and had gaps in the logic, and publication was delayed for years. It is now fundamental to physics and engineering. Should incomplete reasoning ever be accepted if the results work?
Is nature made of sine waves, or is that just our mathematics? If any oscillation can be described as a sum of sine waves, does that mean it really is made of them?
Can a definition be wrong? The second is defined by an atomic oscillation. If we define time this way, what would it even mean to say an atomic clock is "inaccurate"?

How do physics, NoS and ToK fit together? →

1. Describing oscillations

An oscillation is a repeated motion back and forth about an equilibrium position: the position where the object would rest if undisturbed.

A mass on a spring oscillates between +x₀ and −x₀ about the equilibrium position.

Displacement $x$: the distance and direction from equilibrium. It can be positive or negative.
Amplitude $x_0$: the maximum displacement. It is half the distance between the two extremes, not the whole distance.
Time period $T$: the time for one complete oscillation (there and back).
Frequency $f$: the number of oscillations per second, in hertz (Hz).
Angular frequency $\omega$: $2\pi$ radians per cycle, in rad s$^{-1}$. It links SHM to circular motion (A.2).

$$T = \frac{1}{f} = \frac{2\pi}{\omega}$$

Example: a child's swing with a period of 3.2 s has $f = 1/3.2 = 0.31$ Hz and $\omega = 2\pi/3.2 = 2.0\ \text{rad s}^{-1}$.

2. The conditions for SHM

An object moves with simple harmonic motion when the resultant force on it (and so its acceleration):

is proportional to its displacement from the equilibrium position, and
always points towards the equilibrium position, opposite to the displacement.

A force like this is called a restoring force. Written as an equation, this is the defining equation of SHM:

$$a = -\omega^2 x$$

The minus sign matters. It shows that the acceleration is always opposite to the displacement. When the object is to the right of equilibrium ($x$ positive), it accelerates to the left, and vice versa. That's what keeps pulling it back. A graph of $a$ against $x$ is a straight line through the origin with gradient $-\omega^2$.

For SHM, $a$ against $x$ is a straight line through the origin with gradient $-\omega^2$.

Why a mass on a spring does SHM: Hooke's law gives $F = -kx$, and Newton's second law gives $F = ma$. So $ma = -kx$, which means $a = -\frac{k}{m}x$. This has exactly the form $a = -\omega^2 x$, with $\omega^2 = \frac{k}{m}$.

A consequence of SHM is that the period doesn't depend on the amplitude: the motion is isochronous. A bigger swing has further to go, but it moves faster.

3. Graphs of SHM

If you attach a pen to an oscillating mass and pull paper past it at a steady speed, it draws a perfect sine-shaped curve. The displacement, velocity and acceleration all vary sinusoidally with the same period, but they are out of step with each other.

Displacement, velocity and acceleration (black) for a mass released from rest at $+x_0$. The dashed lines mark a quarter and a half of a cycle.

At the extremes ($x = \pm x_0$): the velocity is zero, and the acceleration is at its maximum (pointing back towards equilibrium).
At equilibrium ($x = 0$): the speed is at its maximum and the acceleration is zero.
The acceleration graph is the displacement graph turned upside down. That's the minus sign in $a = -\omega^2 x$.
The velocity is the gradient of the displacement–time graph. The acceleration is the gradient of the velocity–time graph, just as in A.1.

4. The mass–spring system and the simple pendulum

Mass–spring system

Using $\omega^2 = \frac{k}{m}$ and $T = \frac{2\pi}{\omega}$:

$$T = 2\pi\sqrt{\frac{m}{k}}$$

A heavier mass oscillates more slowly, and a stiffer spring (larger $k$) more quickly. The period doesn't depend on the amplitude, or on $g$: the same spring and mass would oscillate at the same rate on the Moon.

Worked example: a mass on a spring

A 0.40 kg mass hangs on a spring with $k = 25\ \text{N m}^{-1}$ and is set oscillating. Find the period and the frequency. What happens to the period if the mass is quadrupled?

$T = 2\pi\sqrt{\dfrac{0.40}{25}} = 2\pi \times 0.126 = 0.79$ s, so $f = \dfrac{1}{0.79} = 1.3$ Hz.
$T \propto \sqrt{m}$, so four times the mass gives $\sqrt{4} = 2$ times the period: 1.6 s.

Simple pendulum

A small mass (the bob) on a light string of length $l$ does SHM, but only for small swings (less than about 10°). For larger angles the restoring force is no longer proportional to the displacement.

$$T = 2\pi\sqrt{\frac{l}{g}}$$

The period depends only on the length and on $g$, not on the mass of the bob or the (small) amplitude.

Worked example: a pendulum clock

(a) Find the period of a pendulum 0.80 m long. (b) What length gives a period of exactly 2.0 s, one second for each swing across?

(a) $T = 2\pi\sqrt{\dfrac{0.80}{9.8}} = 1.8$ s.
(b) Rearranging, $l = g\left(\dfrac{T}{2\pi}\right)^2 = 9.8 \times \left(\dfrac{2.0}{2\pi}\right)^2 = 0.99$ m. Almost exactly a metre.

Measuring g with a pendulum: time 20 oscillations and divide by 20, to reduce the effect of reaction time. Repeat for several lengths and plot $T^2$ against $l$. Squaring $T = 2\pi\sqrt{l/g}$ gives $T^2 = \frac{4\pi^2}{g}l$, a straight line through the origin with gradient $\frac{4\pi^2}{g}$.

5. Energy in SHM

As an oscillator moves, energy is continually transferred between kinetic energy and potential energy: elastic for a spring, gravitational for a pendulum.

At the extremes: the object is momentarily at rest, so $E_k = 0$ and the potential energy is at its maximum.
At equilibrium: the speed is greatest, so $E_k$ is at its maximum and the potential energy is at its minimum.
With no friction or air resistance (no damping), the total energy stays constant.
During one period, $E_k$ and $E_p$ each reach their maximum twice, once on each side. So the energies vary with time at twice the frequency of the oscillation.

Kinetic energy is greatest at the centre, potential energy at the extremes, and the total (dashed) is constant.

At SL you need to describe these energy changes in words and sketches. HL students also calculate them (section 7).

6. Phase and the equations of SHM HL

Imagine a ball moving round a circle at constant angular velocity $\omega$, lit from the side so that its shadow falls on a wall. The shadow moves back and forth in SHM. The ball's angle at any moment, $\omega t$ plus its starting angle, tells you where the shadow is in its cycle. That angle is called the phase.

A point moving in a circle at constant $\omega$. Its projection on a line moves in SHM with displacement $x = x_0\cos\omega t$ (measuring the angle from the horizontal).

The data booklet gives the general equations, where $\phi$ is the phase angle at $t = 0$ (in radians):

$$x = x_0\sin(\omega t + \phi)$$ $$v = \omega x_0\cos(\omega t + \phi)$$ $$v = \pm\,\omega\sqrt{x_0^2 - x^2}$$

$\phi = 0$: the motion starts at equilibrium, moving in the positive direction.
$\phi = \frac{\pi}{2}$: it starts at $+x_0$ (since $\sin(\omega t + \frac{\pi}{2}) = \cos\omega t$).
The maximum speed is $v_{\max} = \omega x_0$, at $x = 0$.

Maths link: deriving the SHM equations by differentiation AA SL 5.6, 5.7 HL

With calculus, all of these equations come from the displacement equation. Velocity is $\dfrac{dx}{dt}$. Use the derivative of $\sin$ and the chain rule: the inside function $\omega t + \phi$ has derivative $\omega$.

$$x = x_0\sin(\omega t + \phi)$$ $$v = \frac{dx}{dt} = \omega x_0\cos(\omega t + \phi)$$

Differentiate again. The derivative of $\cos$ is $-\sin$, and the chain rule gives another factor of $\omega$:

$$a = \frac{dv}{dt} = -\omega^2 x_0\sin(\omega t + \phi)$$

The bracket $x_0\sin(\omega t + \phi)$ is just $x$, so $a = -\omega^2 x$. That's the defining equation of SHM, so this displacement equation really does describe SHM.

The third equation uses the identity $\cos^2\theta + \sin^2\theta = 1$ (AA SL 3.6). Write $\theta = \omega t + \phi$. Then $\cos\theta = \pm\sqrt{1 - \sin^2\theta}$, and $\sin\theta = \dfrac{x}{x_0}$, so:

$$v = \pm\,\omega x_0\sqrt{1 - \frac{x^2}{x_0^2}}$$ $$v = \pm\,\omega\sqrt{x_0^2 - x^2}$$

Two identical oscillators a quarter of a cycle apart have a phase difference of $\frac{\pi}{2}$; half a cycle apart, $\pi$. A phase difference $\Delta\phi$ corresponds to a time difference $\Delta t = \frac{\Delta\phi}{2\pi}T$.

Set your calculator to radians for these equations. It's the most common source of wrong answers.

Worked example: using the SHM equations HL

An object oscillates with amplitude 0.050 m and frequency 2.0 Hz. At $t = 0$ it passes through equilibrium moving in the positive direction. Find its displacement and velocity at $t = 0.10$ s, and its maximum speed.

$x = 0.050\sin(4\pi t)$ over one period ($T = 0.50$ s). At t = 0.10 s the object is still moving outwards, just before it reaches the amplitude at $t = T/4 = 0.125$ s.

$\omega = 2\pi f = 4\pi = 12.6\ \text{rad s}^{-1}$ and $\phi = 0$, so $x = 0.050\sin(4\pi t)$.
At $t = 0.10$ s: $x = 0.050\sin(1.26\ \text{rad}) = 0.048$ m.
$v = \omega x_0\cos(\omega t) = 12.6 \times 0.050 \times \cos(1.26) = 0.19\ \text{m s}^{-1}$.
Check with $v = \omega\sqrt{x_0^2 - x^2} = 12.6\sqrt{0.050^2 - 0.0476^2} = 0.19\ \text{m s}^{-1}$ ✓. $v_{\max} = \omega x_0 = 0.63\ \text{m s}^{-1}$.

7. Energy in SHM: calculations HL

Combining $E_k = \frac{1}{2}mv^2$ with $v = \omega\sqrt{x_0^2 - x^2}$ gives the energies at any displacement:

$$E_T = \tfrac{1}{2}m\omega^2 x_0^2 \qquad E_p = \tfrac{1}{2}m\omega^2 x^2$$ $$E_k = E_T - E_p = \tfrac{1}{2}m\omega^2(x_0^2 - x^2)$$

The total energy is proportional to the square of the amplitude: doubling the amplitude stores four times the energy. For a spring, $m\omega^2 = k$, so $E_p = \frac{1}{2}kx^2$, the same elastic energy as in A.3.

Worked example: energy of an oscillator HL

A 0.20 kg mass oscillates with amplitude 0.050 m and $\omega = 12.6\ \text{rad s}^{-1}$. Find the total energy, and the kinetic and potential energy when $x = 0.025$ m.

$E_T = \tfrac{1}{2}(0.20)(12.6)^2(0.050)^2 = 0.040$ J.
At half the amplitude: $E_p = \tfrac{1}{2}(0.20)(12.6)^2(0.025)^2 = 0.010$ J, which is a quarter of the total, because $E_p \propto x^2$.
So $E_k = 0.040 - 0.010 = 0.030$ J. At half the amplitude, three-quarters of the energy is still kinetic.

End of the HL-only content for C.1.

8. Common mistakes

Confusing the amplitude with the distance between the extremes, which is $2x_0$.
Forgetting the minus sign in $a = -\omega^2 x$, or not being able to explain it.
Saying the velocity is greatest at the extremes. It's zero there. The acceleration is greatest there.
Thinking a heavier pendulum bob swings more slowly. The period of a pendulum doesn't depend on its mass. For a mass–spring system it does.
Using $T = 2\pi\sqrt{l/g}$ for large swings. It only holds for small angles.
HL Calculator in degrees when using $\sin(\omega t + \phi)$. Use radians.
HL Assuming $E_k$ and $E_p$ are equal at half the amplitude. At $x = x_0/2$, $E_p$ is only a quarter of the total.

9. Check your understanding

An astronaut takes a pendulum clock and a mass–spring clock to the Moon. Which one keeps the right time?

The mass–spring clock. Its period, $2\pi\sqrt{m/k}$, doesn't involve $g$. The pendulum's period, $2\pi\sqrt{l/g}$, gets longer where $g$ is smaller, so the pendulum clock runs slow on the Moon.

Where in its oscillation is a mass on a spring when its acceleration is zero?

At the equilibrium position, $x = 0$, since $a = -\omega^2 x$. That's also where its speed is greatest.

Why must the defining equation of SHM have a minus sign?

The restoring force, and so the acceleration, must point back towards equilibrium, opposite to the displacement. Without the minus sign, the acceleration would push the object further away, and it would never return.

A pendulum's length is increased by a factor of 4. What happens to its period?

$T \propto \sqrt{l}$, so the period doubles.

HL An oscillator's amplitude is halved. What happens to its total energy and its maximum speed?

$E_T \propto x_0^2$, so the energy falls to a quarter. $v_{\max} = \omega x_0$, so the maximum speed halves (the period is unchanged).

Practise C.1 questions