Key points
- Simple harmonic motion (SHM) means a = −ω²x: acceleration is proportional to displacement and directed towards equilibrium.
- x = A cos ωt if you start at maximum displacement, x = A sin ωt if you start at the equilibrium position.
- v = ±ω√(A² − x²), so vmax = ωA at x = 0 and amax = ω²A at x = ±A.
- The period T = 2π/ω is independent of amplitude, which is why a pendulum clock keeps time.
- Energy swaps between kinetic and potential, and the total ½mω²A² stays constant if there is no damping.
- Resonance happens when the driving frequency matches the natural frequency, and damping reduces the peak amplitude.
What simple harmonic motion is
Simple harmonic motion, usually shortened to SHM, is the most important type of oscillation in A-level Physics. An object moves in SHM when two things are true: its acceleration is proportional to its displacement from the equilibrium position, and that acceleration is always directed towards the equilibrium position.
In symbols, this defining condition is written as a = −ω²x. The minus sign is the physics: when the displacement is to the right, the acceleration points to the left, and vice versa. The constant ω² just links the size of the acceleration to the size of the displacement. Because acceleration is the same as resultant force divided by mass, SHM needs a restoring force that is proportional to displacement and always pulls the object back, F = −kx for a spring.
If you only remember one thing, remember the test for SHM: a graph of acceleration against displacement is a straight line through the origin with a negative gradient. The gradient is −ω². In an exam, if you are asked to show that motion is simple harmonic, you need to show exactly this: acceleration proportional to displacement, and in the opposite direction.
- Amplitude A: the maximum displacement from equilibrium, in metres.
- Period T: the time for one complete oscillation, in seconds.
- Frequency f: oscillations per second, f = 1/T, in hertz.
- Angular frequency ω: ω = 2πf = 2π/T, in radians per second.
The SHM equations and what each symbol means
These are the equations you need. The OCR A data and formulae booklet gives them, but you must know which one to use and when.
a = −ω²x
- a = acceleration in m s⁻²
- ω = angular frequency in rad s⁻¹
- x = displacement from equilibrium in m
The defining equation of SHM.
x = A cos ωt or x = A sin ωt
- A = amplitude in m
- t = time in s
Use cosine if the oscillation starts at maximum displacement. Use sine if it starts at x = 0.
v = ±ω√(A² − x²)
- v = speed at displacement x, in m s⁻¹
The ± reminds you that the object passes each point in both directions.
vmax = ωA amax = ω²A
The maximum speed occurs at x = 0. The maximum acceleration occurs at x = ±A.
T = 2π/ω = 1/f
Ek = ½mω²(A² − x²) Ep = ½mω²x² E = ½mω²A²
- m = mass in kg
Notice that every equation contains ω. In SHM questions, the first job is nearly always to find ω from the period or frequency, using ω = 2π/T or ω = 2πf. Make sure your calculator is in radian mode whenever you put ωt inside a sine or cosine. This is the most common source of lost marks.
Where the equations come from
You are not asked to derive the SHM equations from scratch, but understanding the link between them makes them far easier to remember. Start from the defining equation a = −ω²x. Velocity is the rate of change of displacement and acceleration is the rate of change of velocity, so a = d²x/dt². Any function whose second derivative is a negative constant times itself is a solution, and sine and cosine have exactly this property.
Try x = A cos ωt. Differentiating gives v = dx/dt = −ωA sin ωt. Differentiating again gives a = dv/dt = −ω²A cos ωt, which is −ω² multiplied by x. So x = A cos ωt satisfies a = −ω²x. The same works for x = A sin ωt. The gradient of the displacement-time graph is the velocity, and the gradient of the velocity-time graph is the acceleration, which is why the graphs below are linked.
The speed formula comes from the identity cos²ωt + sin²ωt = 1. Starting with x = A cos ωt and v = −ωA sin ωt, write cos ωt = x/A and sin ωt = |v|/(ωA). Substitute into the identity: x²/A² + v²/(ω²A²) = 1. Rearranging gives v² = ω²(A² − x²), so v = ±ω√(A² − x²). At x = 0 this gives vmax = ωA, and at x = ±A it gives v = 0, as it must at the turning points.
SHM graphs and phase relationships
Examiners love graphs of displacement, velocity and acceleration against time, because they test whether you understand gradients. The diagram shows all three for an oscillation that starts at maximum positive displacement.
Here is how to read them. The gradient of the displacement-time graph at any point is the velocity. At the peaks the gradient is zero, so v = 0. As the curve crosses the time axis it is steepest, so the speed is at its maximum. The gradient of the velocity-time graph gives the acceleration in the same way.
The phase relationships follow from this. Velocity is a quarter of a cycle (π/2 rad) ahead of displacement: it reaches its maximum a quarter of a period before the displacement does. Acceleration is in antiphase with displacement (a phase difference of π rad, or half a cycle): when x is at its maximum positive value, a is at its maximum negative value. If you are asked to sketch the graph of acceleration, just draw the displacement graph upside down and rescale the axis.
| Position | Displacement | Speed | Acceleration |
|---|---|---|---|
| Equilibrium (x = 0) | 0 | Maximum, ωA | 0 |
| Maximum displacement (x = ±A) | ±A | 0 | Maximum, ω²A, towards equilibrium |
| Halfway (x = A/2) | A/2 | (√3/2)ωA | ω²A/2 |
Energy in SHM
In SHM, energy is continually exchanged between kinetic energy and potential energy. For a mass on a spring the potential energy is stored in the spring. For a pendulum it is gravitational potential energy. At the equilibrium position all the energy is kinetic. At the turning points all of it is potential, and the object is instantaneously at rest.
Because the system is isolated and there is no damping, the total energy is constant: E = ½mω²A². Both Ek and Ep depend on the square of something, so neither goes negative. Plotted against displacement, they are parabolas, and they always add up to the same flat line. If you plot energy against time instead, each of Ek and Ep oscillates at twice the frequency of the motion, because each reaches its maximum twice per cycle.
This gives you a quick check. At x = A/2, Ep = ½mω²(A/2)² = ¼E, so Ek = ¾E. At x = A/√2, the energy is shared equally. Doubling the amplitude does not double the energy: it quadruples it, because E ∝ A².
Two standard examples: the mass-spring system and the simple pendulum
For a mass m on a spring of stiffness k, the restoring force is F = −kx. Using F = ma gives a = −(k/m)x, which matches a = −ω²x with ω² = k/m. The period is therefore T = 2π√(m/k). A stiffer spring gives a shorter period, and a larger mass gives a longer one. The same formula works for a vertical spring, because the extra extension from the weight only shifts the equilibrium position.
T = 2π√(m/k) (mass on a spring)
For a simple pendulum of length L, the restoring force is the component of the weight along the arc, −mg sin θ. For small angles (below about 10°), sin θ ≈ θ in radians, and the displacement along the arc is x = Lθ. This gives a = −(g/L)x, so ω² = g/L and the period is T = 2π√(L/g).
T = 2π√(L/g) (simple pendulum, small angles)
Notice what is missing: the pendulum period does not depend on the mass, and neither pendulum nor spring period depends on the amplitude. That is the key feature of SHM, and it is why a pendulum clock keeps good time as it slowly loses energy. If you plot T² against L for a pendulum, the graph is a straight line through the origin with gradient 4π²/g, which gives a good way to find g from your own measurements. For larger angles the motion stops being simple harmonic, because sin θ is no longer close to θ, and the period starts to rise with amplitude.
Damping, free and forced oscillations, and resonance
A free oscillation happens when a system is displaced and then left to oscillate at its natural frequency f₀. In real systems energy is transferred to the surroundings by friction or air resistance, and the amplitude falls. This is called damping. Light damping gives oscillations whose amplitude decays exponentially while the period stays almost the same. Critical damping returns the system to equilibrium in the shortest time without oscillating. Heavy damping (overdamping) returns it more slowly, again without oscillating. Car suspension and door closers are designed to be close to critical damping.
A forced oscillation happens when a periodic driving force is applied. After the start-up, the system oscillates at the frequency of the driver, not its own natural frequency. As the driving frequency approaches the natural frequency, the amplitude rises sharply. This is resonance. At resonance, the driver transfers energy to the oscillator most efficiently, because the driving force is in step with the velocity.
The graph shows how damping changes the response. Light damping gives a tall, sharp resonance peak. Heavier damping lowers and broadens it, and moves it slightly below the natural frequency. When you answer a question on resonance, say three things: the driving frequency equals the natural frequency, the amplitude is a maximum, and damping reduces the maximum amplitude. Examples to know include a child on a swing, a bridge or building excited by wind or footsteps, and the tuning of a radio circuit. Resonance can be useful or dangerous, so be ready to give one of each and explain how damping helps in the dangerous case.
Worked examples
Mass on a spring: period, maximum speed and maximum acceleration
A mass of 0.40 kg is attached to a spring of stiffness 25 N m⁻¹ and pulled 0.060 m from equilibrium before being released. Calculate (a) the period, (b) the maximum speed and (c) the maximum acceleration.
- (a) ω² = k/m = 25 ÷ 0.40 = 62.5, so ω = 7.906 rad s⁻¹ (1 mark).
- T = 2π/ω = 2π ÷ 7.906 = 0.795 s (1 mark).
- (b) vmax = ωA = 7.906 × 0.060 = 0.474 m s⁻¹ (1 mark).
- (c) amax = ω²A = 62.5 × 0.060 = 3.75 m s⁻² (1 mark).
- Check: the maximum restoring force is kA = 25 × 0.060 = 1.5 N, and 1.5 ÷ 0.40 = 3.75 m s⁻², which agrees (1 mark for a consistent method).
Answer (a) T = 0.80 s (2 s.f.), (b) vmax = 0.47 m s⁻¹, (c) amax = 3.8 m s⁻².
Speed at a given displacement
An object oscillates in SHM with amplitude 0.050 m and frequency 2.0 Hz. Calculate its speed and the magnitude of its acceleration when its displacement is 0.030 m.
- ω = 2πf = 2π × 2.0 = 12.57 rad s⁻¹ (1 mark).
- v = ω√(A² − x²) = 12.57 × √(0.050² − 0.030²) (1 mark).
- A² − x² = 0.0025 − 0.0009 = 0.0016, so the square root is 0.040 m, giving v = 12.57 × 0.040 = 0.503 m s⁻¹ (1 mark).
- a = ω²x = 12.57² × 0.030 = 157.9 × 0.030 = 4.74 m s⁻² (1 mark).
Answer v = 0.50 m s⁻¹ and |a| = 4.7 m s⁻², directed towards the equilibrium position.
Energy in SHM
A 0.50 kg mass oscillates in SHM with period 1.2 s and amplitude 0.040 m. Calculate (a) the total energy and (b) the kinetic energy when the displacement is 0.020 m.
- (a) ω = 2π/T = 2π ÷ 1.2 = 5.236 rad s⁻¹, so ω² = 27.42 rad² s⁻² (1 mark).
- E = ½mω²A² = 0.5 × 0.50 × 27.42 × 0.040² (1 mark).
- E = 0.5 × 0.50 × 27.42 × 0.0016 = 0.01097 J (1 mark).
- (b) Ek = ½mω²(A² − x²) = 0.5 × 0.50 × 27.42 × (0.0016 − 0.0004) (1 mark).
- Ek = 6.855 × 0.0012 = 0.00822 J. This is 75% of E, as expected for x = A/2 (1 mark).
Answer (a) E = 0.011 J, (b) Ek = 0.0082 J.
Using the displacement equation
The displacement of an oscillator is x = 0.050 cos(8.0t), where x is in metres and t is in seconds. Calculate (a) the frequency and (b) the displacement and acceleration at t = 0.20 s.
- (a) Comparing with x = A cos ωt, ω = 8.0 rad s⁻¹ (1 mark).
- f = ω/2π = 8.0 ÷ 2π = 1.27 Hz (1 mark).
- (b) ωt = 8.0 × 0.20 = 1.6 rad. Put the calculator in radian mode (1 mark).
- x = 0.050 × cos(1.6) = 0.050 × (−0.0292) = −1.46 × 10⁻³ m (1 mark).
- a = −ω²x = −64 × (−1.46 × 10⁻³) = +0.093 m s⁻² (1 mark).
Answer (a) f = 1.3 Hz, (b) x = −1.5 × 10⁻³ m and a = +0.093 m s⁻². The mass is just past equilibrium on the negative side, so the acceleration is positive, back towards equilibrium.
Common mistakes
- Calculator in degrees. In x = A cos ωt the angle ωt is in radians. Check your mode before you start.
- Using T instead of ω. Writing v = A/T × something instead of v = ωA. Always convert to ω first.
- Forgetting the minus sign. a = −ω²x. A statement that acceleration is proportional to displacement is incomplete without the words and in the opposite direction (or directed towards equilibrium).
- Saying the amplitude changes the period. In SHM the period is independent of amplitude. For a damped oscillation the period stays almost constant while the amplitude falls.
- Mixing up where the maxima are. Speed is a maximum at the equilibrium position, acceleration is a maximum at the ends.
- Treating a large-angle pendulum as SHM. T = 2π√(L/g) only holds for small angles.
- Confusing damping with resonance. Damping removes energy and reduces amplitude. Resonance is about frequency matching.
- Mixing up the energy graphs. Ek is a maximum at x = 0 and Ep is a maximum at x = ±A. The two parabolas are mirror images, not copies.
How OCR A examines simple harmonic motion
Expect a mixture of short calculations, graph questions and extended explanations. Typical command words are show that, calculate, sketch, explain and describe. This topic sits in Module 5, so start from the module hub to see what surrounds it.
- Calculations: finding ω, vmax, amax, speed at a given displacement, or energy at a given point.
- Graphs: sketching or interpreting x-t, v-t, a-t and energy graphs, and reading the period and amplitude from them. Gradients and areas often appear in the follow-up.
- Explanation: why a system is in SHM, how energy changes during an oscillation, and what happens at resonance with different damping.
- Practical skills: investigating a pendulum or a mass on a spring is a typical way to meet the practical skills requirements, for example plotting T² against L, finding g from the gradient and discussing uncertainties and the benefit of timing many oscillations.
- Synoptic links: circular motion (the shadow of uniform circular motion is SHM, and ω has the same meaning), gravitational fields through the pendulum and g, and the resonance ideas from stationary waves. Damped oscillation amplitudes often decay exponentially, so there are links to the maths of exponentials as well.
Practice questions
A simple pendulum is to have a period of 2.0 s. Calculate its length. Use g = 9.81 m s⁻².3 marks
Show the answer
Answer: L = 0.99 m
T = 2π√(L/g), so L = gT²/4π² = 9.81 × 2.0² ÷ (4π²) = 39.24 ÷ 39.48 = 0.994 m.
A 0.25 kg mass on a spring oscillates with period 0.80 s. Calculate the spring constant.3 marks
Show the answer
Answer: k = 15 N m⁻¹
T = 2π√(m/k), so k = 4π²m/T² = 4π² × 0.25 ÷ 0.64 = 15.4 N m⁻¹, so 15 N m⁻¹ to 2 s.f.
A mass of 0.10 kg on a spring with k = 40 N m⁻¹ oscillates with amplitude 0.020 m. Calculate the frequency, the maximum speed and the maximum acceleration.5 marks
Show the answer
Answer: f = 3.2 Hz, vmax = 0.40 m s⁻¹, amax = 8.0 m s⁻²
ω = √(k/m) = √400 = 20 rad s⁻¹. f = ω/2π = 3.18 Hz. vmax = ωA = 20 × 0.020 = 0.40 m s⁻¹. amax = ω²A = 400 × 0.020 = 8.0 m s⁻².
An oscillator has amplitude 0.10 m and period 2.0 s and is released from the position x = +0.10 m. Calculate the time taken to first reach x = +0.040 m.4 marks
Show the answer
Answer: t = 0.37 s
x = A cos ωt with ω = 2π/2.0 = π rad s⁻¹. cos πt = 0.040/0.10 = 0.40, so πt = 1.159 rad and t = 1.159/π = 0.369 s.
A particle in SHM has total energy E. State the fraction of E that is kinetic when the displacement is half the amplitude, and show your reasoning.3 marks
Show the answer
Answer: Three quarters
Ep = ½mω²x² = ½mω²(A/2)² = ¼ of ½mω²A², so Ep = E/4. Ek = E − Ep = ¾E.
A footbridge starts to sway strongly when a crowd walks across it in step. Explain why, and how damping could reduce the effect.4 marks
Show the answer
Answer: The walking rate acts as a periodic driving force. When its frequency matches the bridge's natural frequency, resonance occurs, energy is transferred efficiently, and the amplitude becomes large. Adding dampers removes energy from the bridge each cycle, lowering and broadening the resonance peak so the maximum amplitude is smaller.
Four marks: driving force, frequencies match, large amplitude from efficient energy transfer, damping dissipates energy so reduces amplitude.
Next steps
SHM links directly to circular motion, because both use ω and radians, so read circular motion next. The pendulum link to g is covered in gravitational fields, and the Kepler's third law derivation uses the same habit of turning a force law into a period. Find all the Module 5 topics on the module hub, and practise this topic with step-by-step marking, free to start.
Frequently asked questions
What is the condition for simple harmonic motion?
The acceleration must be proportional to the displacement from the equilibrium position and always directed towards it. In symbols, a = −ω²x. A graph of acceleration against displacement is a straight line through the origin with a negative gradient.
Why does the period of a pendulum not depend on amplitude?
For small angles the restoring force is proportional to displacement, so ω² = g/L stays constant whatever the amplitude. A larger swing gives a larger restoring force and a longer path, and the two effects cancel. This only holds for small angles, below about 10°.
What is the difference between x = A sin ωt and x = A cos ωt?
Both describe SHM. They differ only in the starting point. Use cosine when the oscillation starts at maximum displacement, and sine when it starts at the equilibrium position moving in the positive direction.
Where are the speed and acceleration greatest in SHM?
The speed is greatest at the equilibrium position, where vmax = ωA. The acceleration is greatest at the maximum displacement, where amax = ω²A. At the equilibrium position the acceleration is zero, and at the extremes the speed is zero.
What is resonance and why does damping matter?
Resonance happens when the driving frequency equals the natural frequency of the oscillator, and the amplitude becomes a maximum. Damping removes energy from the oscillator, so the resonance peak is lower and wider. Without damping the amplitude could grow until the system fails.
Related guides
Written for the OCR A Level Physics A (H556) specification. Questions are original and written in exam style; they are not taken from OCR papers. Spotted a mistake? Tell us.