Displacement, velocity and acceleration graphs for SHMEdexcel A-Level Physics: Revision notes

For an oscillator released from maximum positive displacement at $t = 0$, the displacement–time graph is a cosine curve:

$x = A\cos\omega t$

It starts at $+A$, crosses zero after a quarter of a period ($T/4$), reaches $-A$ at $T/2$, crosses zero again at $3T/4$ and returns to $+A$ at $T$. The amplitude $A$ is the maximum height of the curve and the period $T$ is the time between successive peaks.

If the oscillator starts from equilibrium instead, the curve is a sine curve; the shape is the same, shifted by a quarter of a cycle.

Velocity is the gradient of the displacement–time graph: $v = \Delta x/\Delta t$ (use a tangent for a curve). Reading the cosine curve:

• at $x = \pm A$ the graph is flat, so $v = 0$
• at $x = 0$ the graph is steepest, so the speed is a maximum, $v_{max} = A\omega$
• from $+A$ to $-A$ the gradient is negative, so $v$ is negative

The velocity–time graph is therefore a negative sine curve:

$v = -A\omega\sin\omega t$

It starts at zero, reaches $-A\omega$ at $T/4$, is zero again at $T/2$, and reaches $+A\omega$ at $3T/4$. Its maximum value is $A\omega$.

Acceleration is the gradient of the velocity–time graph. The velocity graph is flat at its maximum and minimum (so $a = 0$ at $x = 0$) and steepest when it crosses zero (so $|a|$ is largest at $x = \pm A$). The acceleration–time graph is a negative cosine curve:

$a = -A\omega^2\cos\omega t$

Its maximum value is $A\omega^2$. Since $x = A\cos\omega t$, this is $a = -\omega^2 x$, so the acceleration graph is a mirror image of the displacement graph: it is in antiphase.

Worked example: $A = 0.12$ m, $T = 1.6$ s, so $\omega = 3.93$ rad s⁻¹: $v_{max} = 0.47$ m s⁻¹, $a_{max} = 1.85$ m s⁻².

All three graphs have the same period. Their key points over one cycle, starting from $x = +A$:

• $t = 0$: $x = +A$, $v = 0$, $a = -A\omega^2$ (maximum, towards equilibrium)
• $t = T/4$: $x = 0$, $v = -A\omega$ (maximum speed), $a = 0$
• $t = T/2$: $x = -A$, $v = 0$, $a = +A\omega^2$
• $t = 3T/4$: $x = 0$, $v = +A\omega$, $a = 0$

The velocity graph is a quarter of a cycle ahead of the displacement graph, and the acceleration graph is a further quarter of a cycle ahead (half a cycle from displacement). Whenever $|x|$ is largest, $v$ is zero and $|a|$ is largest; whenever $x = 0$, $|v|$ is largest and $a = 0$.

To interpret a graph:

• Amplitude from the maximum value of the displacement graph
• Period from the time between two successive peaks, then $f = 1/T$ and $\omega = 2\pi/T$
• Velocity at a time from the gradient of the tangent to the displacement graph
• Acceleration at a time from the gradient of the tangent to the velocity graph
• Check with the equations: for example $a = -\omega^2 x$ at any time

Worked example: $x = 0.060\cos(\pi t)$ m. At $t = 0.50$ s, $v = -0.060\pi\sin(0.5\pi) = -0.19$ m s⁻¹: the gradient of the displacement graph there is negative, as expected.