Graphs of simple harmonic motionEdexcel International A Level Physics: Revision notes

For an oscillator released from maximum positive displacement, $x = A\cos\omega t$, so the displacement–time graph is a cosine curve with amplitude $A$ and period $T = 2\pi/\omega$.

The gradient of the graph at any point is the rate of change of displacement, which is the velocity at that point:

• At maximum displacement ($x = \pm A$) the gradient is zero, so the object is momentarily at rest.
• At equilibrium ($x = 0$) the gradient has its maximum magnitude, so the speed is greatest.
• The gradient is positive when the object moves in the positive direction and negative when it moves in the negative direction.

The velocity is $v = -A\omega\sin\omega t$, a negative sine curve with maximum value $v_{max} = A\omega$.

It is the gradient of the displacement–time graph at each time:

• The velocity is zero when the displacement is a maximum.
• The velocity is a maximum when the displacement is zero.
• The velocity graph is a quarter of a cycle ahead of the displacement graph.

The gradient of the velocity–time graph at any point is the acceleration at that point.

The acceleration is $a = -A\omega^2\cos\omega t = -\omega^2 x$, with maximum magnitude $a_{max} = A\omega^2$.

• The acceleration is zero when the velocity is a maximum (at equilibrium).
• The acceleration is greatest when the velocity is zero (at the ends of the motion).
• Since $a = -\omega^2 x$, the acceleration graph is the displacement graph inverted: they are half a cycle out of phase (antiphase).

All three graphs have the same period. Each graph is the gradient of the one before it, and each moves a quarter of a cycle ahead.

A trolley has $A = 0.12$ m and $T = 0.80$ s, and is released from maximum positive displacement. Then $\omega = 2\pi/0.80 = 7.85$ rad s⁻¹.

At $t = 0.10$ s, $\omega t = 0.785$ rad, so

$v = -0.12 \times 7.85 \times \sin 0.785 = -0.67$ m s⁻¹

$a = -0.12 \times 7.85^2 \times \cos 0.785 = -5.2$ m s⁻²

Both are negative, so the acceleration is in the direction of the velocity and the trolley is speeding up as it moves towards equilibrium.

The maximum gradients are $A\omega$ (displacement graph) and $A\omega^2$ (velocity graph).