C.1 Simple harmonic motion

An oscillation is a repeated back-and-forth motion about a fixed position. Pendulums, vibrating guitar strings, the cone of a loudspeaker, atoms in a solid and the swaying of a tall building all oscillate. Many of these move in a particular way called simple harmonic motion, which can be described completely by a small set of equations and graphs. The same model later explains how waves are produced and how the particles of a medium move as a wave passes.

Pendulum clocks were the most accurate timekeepers for almost three centuries because the period of a pendulum swinging through a small angle hardly depends on how far it swings. Engineers use the same physics in reverse: the tallest buildings carry heavy pendulums tuned to swing against the building's own sway.

Describing an oscillation

Every oscillating body has an equilibrium position: the position where the resultant force on it is zero and where it would stay at rest if undisturbed. A mass on a spring rests where the spring force balances any other forces; a pendulum bob rests directly below its pivot. When the body is moved away and released, it moves back towards the equilibrium position, overshoots it and continues to the other side, again and again.

The quantities used to describe the motion are measured from the equilibrium position. The displacement $x$ is the distance of the body from equilibrium in a stated direction, so it is positive on one side and negative on the other. The largest displacement reached is the amplitude $x_0$. The symbol $x_0$ for amplitude is the one used in the data booklet.

The diagram of the block on a spring shows one complete cycle: from one extreme, through equilibrium, to the other extreme and back again. A cycle can start at any point, as long as it ends when the body is next at the same position moving in the same direction.

Period, frequency and angular frequency

If one oscillation takes a time $T$, the number of oscillations in one second is $1/T$. The period and frequency are therefore reciprocals of each other. It is also useful to describe an oscillation by its angular frequency $\omega$, measured in radians per second.

The angular frequency comes from the link between SHM and circular motion. A point moving round a circle at constant angular speed $\omega$ turns through $2\pi$ radians in one revolution. Its shadow on a wall moves back and forth in simple harmonic motion with the same period, so one oscillation corresponds to a phase change of $2\pi$ radians. A large $\omega$ means a rapid oscillation.

The conditions for simple harmonic motion

A body oscillates only if a force pulls it back whenever it is displaced. This restoring force always points towards the equilibrium position. Motion is simple harmonic when the restoring force has a particular form: its size is proportional to the displacement.

Here $k$ is a constant for the system, and the minus sign shows that the force acts in the opposite direction to the displacement. Displace the body twice as far and the restoring force doubles. By Newton's second law, $a = F/m = -(k/m)x$, so the acceleration is also proportional to the displacement and opposite to it. Writing the constant $k/m$ as $\omega^2$ gives the defining equation of SHM.

Both conditions matter. A ball bouncing on the floor repeats its motion, but its acceleration is $g$ downwards throughout the flight, whatever its height, so the motion is periodic without being simple harmonic. A body is in simple harmonic motion only when $a$ is proportional to $x$.

The meaning of the minus sign

The minus sign in $a = -\omega^2 x$ does not mean that the body is slowing down. It means that the acceleration always points back towards equilibrium. When the body is on the positive side ($x > 0$), its acceleration is negative; on the negative side it is positive. Whether the body is speeding up or slowing down depends on its direction of motion as well.

• Moving towards equilibrium, the velocity and acceleration are in the same direction, so the body speeds up.

• Moving away from equilibrium, they are in opposite directions, so the body slows down until it stops momentarily at the extreme.

The acceleration–displacement graph

Because $a = -\omega^2 x$, a graph of acceleration against displacement is a straight line through the origin with a negative gradient. The gradient equals $-\omega^2$. The line stops at $x = \pm x_0$, because the body never goes beyond its amplitude. Any oscillation whose $a$–$x$ graph has this shape is simple harmonic.

A force law can be tested in the same way. If the resultant force on a body is written as $F = -(\text{constant})\times x$, dividing by the mass gives $a = -\omega^2 x$ with $\omega^2 = \text{constant}/m$.

Why the period does not depend on the amplitude

In SHM, a body released from a larger displacement has further to travel, but it also experiences a proportionally larger restoring force and so a larger acceleration throughout its motion. The two effects cancel exactly, and the period is the same for any amplitude. Motion with this property is called isochronous. It is the reason a pendulum clock keeps time even as its swing slowly changes, and the reason a musical note keeps its pitch as it fades.

Graphs of displacement, velocity and acceleration

Plotting displacement against time for SHM gives a sinusoidal curve: a sine or cosine shape, depending on where the timing starts. The graphs of velocity and acceleration against time are also sinusoidal, with the same period, but shifted along the time axis. Aligning the three graphs one above the other makes the relationships clear.

Reading one graph from another

The three graphs are linked by gradients, exactly as in kinematics (A.1).

• Velocity is the gradient of the $x$–$t$ graph. At the extremes ($x = \pm x_0$) the $x$–$t$ graph is flat, so $v = 0$. As the body passes through equilibrium the graph is steepest, so the speed is a maximum there.

• Acceleration is the gradient of the $v$–$t$ graph, and from $a = -\omega^2 x$ it is also $-\omega^2$

The velocity graph is a quarter of a cycle ahead of the displacement graph: each feature of the $v$–$t$ curve happens $T/4$ earlier than the matching feature of the $x$–$t$ curve. The acceleration graph is half a cycle away from the displacement graph, so the two are always opposite in sign. Oscillations in this relationship are said to be in antiphase.

The mass–spring system

A mass $m$ on a light spring of spring constant $k$ is the standard example of a mass–spring system. When the mass is displaced by $x$ from equilibrium, the spring exerts a restoring force $F = -kx$ (Hooke's law, from A.2). Comparing $a = -(k/m)x$ with $a = -\omega^2 x$ gives $\omega^2 = k/m$, and so the period.

The equation agrees with experience. A larger mass has more inertia, so it accelerates less for the same force and its period is longer. A stiffer spring (larger $k$) pulls back harder, so its period is shorter. Gravity does not appear: a mass on a spring has the same period on the Moon as on the Earth, and the amplitude does not appear either.

Changing the mass or the spring

Because $T \propto \sqrt{m}$, quadrupling the mass doubles the period. Because $T \propto 1/\sqrt{k}$, a spring four times as stiff halves it. Questions of this kind are answered fastest with ratios rather than by working out each period separately.

A mass hanging on a vertical spring

When a mass hangs from a vertical spring, its weight stretches the spring by an extension $e$ until the spring force balances the weight: $ke = mg$. This is the new equilibrium position. If the mass is pulled down a further distance $x$, the spring force increases by $kx$ while the weight is unchanged, so the resultant force is $-kx$, exactly as for a horizontal spring.

The vertical diagrams show that gravity only moves the equilibrium position. The oscillation about that position is SHM with $T = 2\pi\sqrt{m/k}$. Since $m/k = e/g$, the period can also be found from the static extension alone.

Measuring a spring constant from oscillations

Squaring the period equation gives $T^2 = (4\pi^2/k)\,m$. A graph of $T^2$ against $m$ should therefore be a straight line through the origin with gradient $4\pi^2/k$. Plotting $T^2$ rather than $T$ turns a curve into a straight line, which is easier to test and from which a constant can be found by measuring a gradient.

In a real experiment the line usually has a small positive intercept on the $T^2$ axis. The spring itself has mass, and the lower part of it moves with the load, so the oscillating mass is slightly larger than the mass on the hanger. Using the gradient rather than a single reading removes this systematic effect from the value of $k$.

The simple pendulum

A simple pendulum is a small, dense bob on a light, inextensible string of length $l$, swinging in a vertical plane. The length $l$ is measured from the pivot to the centre of mass of the bob. When the bob is displaced by an angle $\theta$, two forces act on it: the tension $F_\mathrm{T}$ in the string and its weight $mg$.

Resolving the weight along and perpendicular to the string, as in the pendulum diagram, separates the two jobs the forces do. Along the string, $F_\mathrm{T} - mg\cos\theta$ provides the centripetal force that keeps the bob on its circular path. Along the arc, the only force is the component $mg\sin\theta$, and it always points back towards the lowest point. This component is the restoring force.

Why a pendulum moves with SHM only for small angles

The restoring force is $F = -mg\sin\theta$. If the angle is small and measured in radians, $\sin\theta \approx \theta$, and the angle is related to the arc displacement by $\theta = x/l$. Then

This has the form $a = -\omega^2 x$ with $\omega^2 = g/l$, so for small swings the pendulum moves with SHM and its period is

The small-angle approximation $\sin\theta \approx \theta$ is the key step. For larger angles $\sin\theta$ is noticeably smaller than $\theta$, the restoring force grows more slowly than the displacement, and the period becomes slightly longer than the equation predicts. The values of $\sin\theta$ and $\theta$ show how good the approximation is.

Up to about 10° the difference is well under 1%, so the pendulum equation is accurate for the small swings used in timekeeping and in the laboratory.

What the pendulum period depends on

The mass of the bob does not appear in $T = 2\pi\sqrt{l/g}$. A heavier bob needs a larger force to give it the same acceleration, and its weight, which supplies the restoring force, is larger in the same proportion. This is the same reason all bodies in free fall have the same acceleration. The amplitude does not appear either, as long as it is small. Only the length and the gravitational field strength matter.

• $T \propto \sqrt{l}$: four times the length gives twice the period.

• $T \propto 1/\sqrt{g}$: a pendulum swings more slowly on the Moon, where $g$ is about one-sixth of its value on the Earth.

The Great Clock in the Elizabeth Tower at Westminster ("Big Ben") has a pendulum that beats once every 2 seconds, so its period is 4.0 s. The pendulum has a total length of 4.4 m, but it is a heavy rod and bob rather than a simple pendulum, so the length that matters is its effective length, the length of a simple pendulum with the same period: $l = gT^2/4\pi^2 = 9.8 \times 4.0^2 / 4\pi^2 \approx 4.0\ \text{m}$. The clock is fine-tuned by placing old penny coins on a small tray on the pendulum: each coin raises the centre of mass very slightly, shortening the effective length and making the clock gain about 0.4 s per day.

Measuring g with a pendulum

Rearranging the pendulum equation gives $g = 4\pi^2 l/T^2$. Measuring $T$ for several lengths and plotting $T^2$ against $l$ gives a straight line through the origin of gradient $4\pi^2/g$, from which $g$ is found. Good practice in this experiment includes:

• measuring the length to the centre of the bob, using a metre rule held close to the string;

• keeping the amplitude small (under about 10°), so that the small-angle approximation holds;

• timing 20 oscillations from a fiducial marker at the equilibrium position and repeating each timing;

• checking that the bob swings in a plane rather than in an ellipse.

Energy changes during simple harmonic motion

As a body oscillates, energy is transferred back and forth between kinetic energy and potential energy. For a mass on a horizontal spring the potential energy is elastic potential energy stored in the spring; for a pendulum it is gravitational potential energy of the bob. If no energy is lost to the surroundings, the total energy stays constant (energy conservation, A.3), and the amplitude stays the same from one cycle to the next.

• At each extreme the body is momentarily at rest: the kinetic energy is zero and the potential energy is a maximum, equal to the total energy.

• At the equilibrium position the speed is greatest: the kinetic energy is a maximum, equal to the total energy, and the potential energy is a minimum (taken as zero).

• In between, the energy is shared, and the two always add up to the same total.

Energy against time

The kinetic energy is greatest each time the body passes through equilibrium, which happens twice in every cycle (once in each direction). The potential energy likewise peaks twice per cycle, at each extreme. Graphs of energy against time therefore have half the period of the displacement graph, and since energy cannot be negative, they never go below zero.

Energy against displacement

Plotted against displacement, the potential energy is a parabola with its minimum at $x = 0$, because the restoring force, and so the work needed to displace the body, grows with $x$. The kinetic energy is an upside-down parabola that reaches zero at $x = \pm x_0$. The graphs end at the amplitude, because the body never goes further.

Real oscillators always lose some energy, for example through fluid resistance or friction at a pivot, so their amplitude slowly decreases. This decay, called damping, is part of standing waves and resonance (C.4). Taipei 101, a 508 m tower in a region with typhoons and earthquakes, carries a 660-tonne tuned mass damper: a steel sphere hung as a pendulum, with a period matched to the building's sway. As the building sways, the sphere swings against it, and hydraulic dampers attached to the sphere convert the energy of the motion into internal energy.

Phase and phase angle (HL)

One complete cycle of SHM corresponds to $2\pi$ radians, as for one revolution of the point on a circle whose shadow moves in SHM. The phase of an oscillation is its position in the cycle expressed as an angle. At time $t$ an oscillator with angular frequency $\omega$ is at phase $\omega t + \phi$, where $\phi$ is its phase angle at $t = 0$, which sets where the oscillation starts.

The reference circle shows where the phase comes from. A point moving round a circle of radius $x_0$ at angular speed $\omega$ has a height $x_0\sin(\omega t + \phi)$ above the centre, where $\omega t + \phi$ is the angle its radius makes with the horizontal. As the angle increases by $2\pi$, the height runs through one complete cycle of SHM. The phase angle $\phi$ is the angle of the radius at $t = 0$.

Phase difference between two oscillators

Two oscillators with the same frequency keep a constant phase difference $\Delta\phi$. If one reaches each stage of its cycle a time $\Delta t$ after the other, the fraction of a cycle by which it lags is $\Delta t/T$, so

Two oscillators are in phase when their phase difference is zero, and in antiphase when it is $\pi$. In this language, the velocity of an oscillator leads its displacement by $\pi/2$, and its acceleration is in antiphase with its displacement.

The equations of simple harmonic motion (HL)

The sinusoidal graphs of SHM are described by two equations. With $x_0$ the amplitude and $\phi$ the phase angle,

The value of $\phi$ depends on where the timing starts.

• Starting at equilibrium, moving in the positive direction: $\phi = 0$, so $x = x_0\sin\omega t$ and $v = \omega x_0\cos\omega t$.

• Starting at the positive extreme: $\phi = \pi/2$, and since $\sin(\theta + \pi/2) = \cos\theta$, $x = x_0\cos\omega t$

Because $\cos$ is never greater than 1, the maximum speed is $v_\text{max} = \omega x_0$, reached at equilibrium. Similarly the maximum magnitude of acceleration, at the extremes, is $a_\text{max} = \omega^2 x_0$.

Speed at a given displacement

Often the time is not known but the displacement is. Combining the two equations with $\sin^2\theta + \cos^2\theta = 1$ removes the time:

The $\pm$ sign appears because the body passes each point twice per cycle, once in each direction, with the same speed. At $x = 0$ the equation gives $v = \pm\omega x_0$, the maximum speed; at $x = \pm x_0$ it gives $v = 0$.

The equations also show why loudspeakers and engine parts must be strong. A loudspeaker cone vibrating at 250 Hz with an amplitude of only 0.40 mm has $\omega = 1.57\times10^3\ \text{rad s}^{-1}$, so its maximum acceleration is $\omega^2 x_0 \approx 1.0\times10^3\ \text{m s}^{-2}$, about a hundred times $g$.

Energy in simple harmonic motion (HL)

The kinetic energy at equilibrium equals the total energy, because the potential energy is zero there. Using $v_\text{max} = \omega x_0$:

At a displacement $x$, the kinetic energy is $\tfrac12 mv^2 = \tfrac12 m\omega^2(x_0^2 - x^2)$, and the potential energy is the remainder of the total:

For a mass on a spring $m\omega^2 = k$, so $E_\mathrm{p} = \tfrac12 kx^2$, the elastic potential energy from A.3. The SHM equation is more general: it applies to any simple harmonic oscillator, including a pendulum, where the "spring constant" is $m\omega^2 = mg/l$.

• The total energy is proportional to the square of the amplitude: doubling $x_0$ quadruples $E_\mathrm{T}$.

• It is also proportional to $\omega^2$, so for the same mass and amplitude, doubling the frequency quadruples the energy.

Applying the ideas in exam questions

Questions on SHM combine three skills: recognising the defining condition, reading phase relationships from graphs, and choosing the right period equation. Data-based questions on springs and pendulums usually involve a linearised graph and its uncertainties.

Paper 1A: the acceleration of an oscillator

Paper 1A: changing a pendulum

Paper 1B: determining g from a pendulum

A student investigates how the period $T$ of a simple pendulum varies with its length $l$. For each length she times 10 oscillations. The uncertainty in each time for 10 oscillations is $\pm 0.2\ \text{s}$.

Paper 2: a mass on a spring

Paper 2 (HL): speed and energy of an oscillator

Self-check

• To do:

  I can define amplitude, period, frequency and angular frequency, and convert between T, f and ω.

• To do:

  I can state the two conditions for simple harmonic motion and explain the minus sign in a = −ω²x.

• To do:

  I can identify SHM from an a–x graph or a force law and find ω from the gradient.

• To do:

  I can explain why the period of SHM does not depend on the amplitude.

• To do:

  I can sketch and relate the x–t, v–t and a–t graphs, and state where speed and acceleration are greatest.

• To do:

  I can derive and use T = 2π√(m/k) for a mass on a spring, including a vertical spring.

• To do:

  I can explain why a simple pendulum moves with SHM only for small angles and use T = 2π√(l/g).

• To do:

  I can find a spring constant or g from a linearised graph and handle the uncertainties.

• To do:

  I can describe the energy changes during an oscillation and sketch energy against time and against displacement.

• To do:

  I can describe oscillations using phase angle, find the phase angle from a graph or starting conditions, and calculate phase differences in radians (HL).

• To do:

  I can use x = x₀ sin(ωt + φ), v = ωx₀ cos(ωt + φ) and v = ±ω√(x₀² − x²) (HL).

• To do:

  I can calculate the total, potential and kinetic energy of a simple harmonic oscillator (HL).

1. (a)

   Calculate the speed of the block and dart immediately after the collision.

   [2]2 marks

   Markscheme (a)

   $0.020 \times 15 = 0.400\,v$ ✓
   $v = 0.75$ «m s$^{-1}$» ✓
   Award [2] for a bald correct answer.

2. (b)

   Calculate the percentage of the dart's initial kinetic energy that is transferred to other forms in the collision.

   [2]2 marks

   Markscheme (b)

   before: 2.25 «J» AND after: $\frac{1}{2} \times 0.400 \times 0.75^2 = 0.113$

3. (c)

   The block and dart then oscillate with simple harmonic motion. Calculate the period of the oscillations.

   [2]2 marks

   Markscheme (c)

   $T = 2\pi\sqrt{\frac{0.400}{40}}$

4. (d)

   Determine the amplitude of the oscillations.

   [2]2 marks

   Markscheme (d)

   the speed just after the collision is the maximum speed, $v_{\max} = \omega x_0$ ✓
   $x_0 = \frac{0.75}{10} = 0.075$

5. (e)

   Calculate the maximum force exerted on the block by the spring, and the maximum acceleration of the block.

   [2]2 marks

   Markscheme (e)

   $F_{\max} = kx_0 = 40 \times 0.075 = 3.0$

The amplitude, 0.050 m, does not enter the calculation: the period depends only on the gradient of the line.

The unit of the gradient is $\text{s}^2\ \text{kg}^{-1}$, and $\text{kg s}^{-2}$ is the same as $\text{N m}^{-1}$, so the answer has the correct unit for a spring constant.

Use the direction of motion to choose. The velocity at $t = 0$ is $v = \omega x_0\cos\phi$. The displacement is increasing, so $v > 0$ and $\cos\phi > 0$. Of the two angles, only $\pi/6$ has a positive cosine.

If the displacement had been $+2.0\ \text{cm}$ and decreasing, the answer would be $5\pi/6$. On the reference circle the two cases are the two points at the same height, one on each side of the vertical diameter.

percentage uncertainty in $T^2 = 2 \times 2.2 = 4.4\%$ «accept 4%» ✓

(c) $T^2 = \dfrac{4\pi^2}{g}\,l$, so $T^2$ is proportional to $l$ ✓

a straight line through the origin is expected, whose gradient $4\pi^2/g$ gives $g$ ✓

(d) gradient $= \dfrac{4.04 - 0.81}{1.000 - 0.200} = 4.04$ «s² m⁻¹» ✓

$g = \dfrac{4\pi^2}{4.04} = 9.8$ «m s⁻²» ✓

Allow ECF from an incorrect gradient. Accept 9.7–9.9.

(e) reduces the percentage uncertainty in $T$ «as the reaction-time uncertainty is shared over 10 periods» ✓