SHM Solutions — Displacement, Velocity and Acceleration

In the previous lesson we defined simple harmonic motion by the equation a = −ω²x. That is a differential equation in disguise: if we write a = d²x/dt², it becomes

d²x/dt² = −ω² x

Solving this equation gives us explicit formulae for the displacement, velocity and acceleration of an oscillating body as functions of time. These are the "SHM solutions" you need to use confidently in OCR A-Level Physics A Module 5.3.

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The General Solution

Any function whose second derivative is equal to the original function multiplied by −ω² must be a sine or cosine. The two most useful forms are:

x(t) = A cos(ωt) — if the oscillator starts at x = +A when t = 0 x(t) = A sin(ωt) — if the oscillator starts at x = 0 when t = 0

A is the amplitude — the maximum displacement from equilibrium. ω is the angular frequency we met in the last lesson.

Both are solutions of a = −ω²x and both describe genuine SHM. Which one you use depends on where the oscillator was at t = 0. The OCR exam tends to favour the cosine form, so that is what we will concentrate on here.

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Velocity of an SHM Oscillator

If x = A cos(ωt), then by differentiation

v = dx/dt = −Aω sin(ωt)

This tells us two things immediately:

1. The velocity is itself sinusoidal, with the same angular frequency ω but shifted by a quarter cycle (the sine function is 90° behind cosine).
2. The maximum speed is v_max = Aω, which occurs whenever sin(ωt) = ±1, i.e. when cos(ωt) = 0, i.e. when the oscillator passes through equilibrium.

At the extremes of the motion, where x = ±A, the velocity is zero — the oscillator momentarily stops before being pulled back.

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Acceleration of an SHM Oscillator

Differentiating again:

a = dv/dt = −Aω² cos(ωt) = −ω²x

which is consistent with the defining equation, as it must be. The maximum acceleration has magnitude

a_max = Aω²

and occurs at the extremes of the motion, exactly where the velocity is zero. At equilibrium (x = 0) the acceleration is also zero — makes sense, because there is no net force at equilibrium.

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Summary of Maxima

Quantity	Formula	When does it occur?
Maximum displacement	A	At the extremes of the motion
Maximum speed	Aω	Passing through equilibrium (x = 0)
Maximum acceleration	Aω²	At the extremes of the motion (x = ±A)

Memorising this table will save you time under exam conditions.

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The Velocity–Displacement Relationship

It is often more useful to know the speed at a given position, not at a given time. We can eliminate t between x = A cos(ωt) and v = −Aω sin(ωt) using the Pythagorean identity sin²(ωt) + cos²(ωt) = 1:

sin²(ωt) = 1 − cos²(ωt) = 1 − (x/A)²

So

v² = A²ω² sin²(ωt) = A²ω² (1 − x²/A²) = ω² (A² − x²)

Taking the square root:

v = ±ω√(A² − x²)

The ± sign reminds you that the oscillator passes through each position twice per cycle — once moving left, once moving right.

• At x = 0, v = ±ωA (maximum speed).
• At x = ±A, v = 0 (momentarily stationary).

This equation appears constantly in OCR exam papers and is the single most useful equation in the whole SHM toolkit.

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Phase and Phase Difference

Two SHM oscillators with the same angular frequency may still be out of step with each other. The difference is called the phase difference, usually written φ and measured in radians.

• x₁ = A cos(ωt) and x₂ = A cos(ωt) are in phase — phase difference 0.
• x₁ = A cos(ωt) and x₂ = A cos(ωt + π) are in antiphase — phase difference π (one is at +A when the other is at −A).
• x₁ = A cos(ωt) and x₂ = A sin(ωt) = A cos(ωt − π/2) are π/2 out of phase (a quarter cycle).

The displacement and velocity of a single oscillator are π/2 out of phase (sine leads cosine by a quarter cycle). The displacement and acceleration are π out of phase (in antiphase) — hence the minus sign in a = −ω²x.

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The SHM Cycle — a Visual Summary

Picture a mass on a spring, released from rest at x = +A. Over one period T: