SHM Equations: Displacement, Velocity & Time Revision Notes | A Level AQA Physics

1. Why These Equations Matter

You already know that SHM is defined by $a = -\omega^2 x$. That tells you the relationship between acceleration and displacement — but it says nothing about where the object is at a specific moment in time, or how fast it is moving. That is exactly what this lesson fills in.

The two equations $x = A\cos(\omega t)$ and $v = \pm\omega\sqrt{A^2 - x^2}$ are the workhorses of SHM calculations at A Level. Master these and you can answer almost any AQA question on oscillations — from finding the speed at a given position to identifying the instant of maximum acceleration.

2. Displacement as a Function of Time

Picture a mass on a spring pulled to its maximum displacement $A$ and released from rest at $t = 0$. At that instant the displacement equals $A$, and as time progresses the mass oscillates back and forth. The displacement at any later time is given by:

The cosine function starts at its maximum value of 1 when $t = 0$, so $x = A$ at the start — exactly matching the physical setup. As $\omega t$ increases, the cosine oscillates between $+1$ and $-1$, so displacement oscillates between $+A$ and $-A$. The angular frequency $\omega$ controls how quickly those oscillations happen.

Reading the graph: at $t = 0$ the object is at $+A$; at $t = T/4$ it passes through equilibrium; at $t = T/2$ it reaches $-A$; and the cycle repeats. This cosine shape is the fingerprint of SHM starting from maximum displacement.

3. Velocity as a Function of Displacement

Knowing where the object is at time $t$ is useful, but examiners frequently ask for the speed at a particular position rather than at a particular time. For that, a second equation links velocity directly to displacement:

Notice what happens at the two extremes. When $x = A$ (the turning point), the square root becomes $\sqrt{A^2 - A^2} = 0$, so $v = 0$ — the object momentarily stops. When $x = 0$ (the equilibrium position), the square root becomes $\sqrt{A^2} = A$, giving $v = \pm\omega A$. That is the maximum speed, and it occurs every time the object passes through equilibrium.

This result — $v_{max} = \omega A$ — is worth committing to memory as a standalone formula. It tells you that doubling the amplitude doubles the maximum speed, and doubling the frequency (which doubles $\omega$) also doubles the maximum speed. Both factors matter equally.

4. Maximum Acceleration

Recall from the defining equation of SHM that $a = -\omega^2 x$. The magnitude of acceleration is largest when the magnitude of displacement is largest — that is, at the turning points where $|x| = A$.

There is a striking contrast here: maximum speed and maximum acceleration occur at opposite positions. Maximum speed is at $x = 0$; maximum acceleration is at $x = \pm A$. This is because the restoring force — and therefore the acceleration — is proportional to displacement, so it is greatest where displacement is greatest.

5. Worked Calculations

Theory is only useful once you can apply it under exam conditions. The following worked examples cover the three most common question types: finding displacement at a given time, finding speed at a given position, and finding maximum values. Work through each step before revealing the next.

Worked example: Finding displacement at a given time

A mass on a spring oscillates with amplitude $0.080$ m and period $2.0$ s. It is released from maximum displacement at $t = 0$. Calculate the displacement at $t = 0.50$ s. [3 marks]

1. $\omega = \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi$ rad s$^{-1}$ — [M1] Always find $\omega$ first. Here $\omega = \pi \approx 3.14$ rad s$^{-1}$.
2. $x = A\cos(\omega t)$ — Released from maximum displacement at $t = 0$, so cosine form applies.
3. $x = 0.080 \times \cos(\pi \times 0.50)$ — [M1] The argument is $\omega t = \pi \times 0.50 = 0.50\pi$ radians.
4. $\cos(0.50\pi) = \cos(90°) = 0$ — $0.50\pi$ radians equals $90°$. Cosine of $90°$ is exactly $0$.
5. $x = 0.080 \times 0 = 0$ m — [A1] At $t = 0.50$ s (which is $T/4$), the mass is passing through the equilibrium position. This makes physical sense.

Answer: $x = 0$ m (the mass is at the equilibrium position)

Worked example: Finding speed at a given displacement

A pendulum oscillates in SHM with amplitude $A = 0.15$ m and angular frequency $\omega = 4.0$ rad s$^{-1}$. Calculate the speed of the pendulum when its displacement is $x = 0.090$ m. [3 marks]

1. $v = \pm\omega\sqrt{A^2 - x^2}$ — [M1] This equation links speed directly to position — no need to find time first.
2. $A^2 = 0.15 \times 0.15 = 0.0225$ m$^2$ $x^2 = 0.090 \times 0.090 = 0.0081$ m$^2$ — Expanding each square longhand avoids arithmetic errors. Note: $0.090^2 eq 0.081$.
3. $A^2 - x^2 = 0.0225 - 0.0081 = 0.0144$ m$^2$ — [M1] Subtraction inside the square root.
4. $\sqrt{0.0144} = 0.12$ m — Check: $0.12 \times 0.12 = 0.0144$. Confirmed.
5. $v = \pm 4.0 \times 0.12 = \pm 0.48$ m s$^{-1}$ — [A1] The $\pm$ sign indicates the pendulum could be moving in either direction at this displacement. The speed (magnitude) is $0.48$ m s$^{-1}$.

Answer: $|v| = 0.48$ m s$^{-1}$

Worked example: Finding maximum speed and maximum acceleration

An object oscillates in SHM with amplitude $0.050$ m and period $0.40$ s. Calculate (a) the maximum speed and (b) the maximum acceleration. [4 marks]

1. $\omega = \frac{2\pi}{T} = \frac{2\pi}{0.40}$ $\omega = \frac{6.2832}{0.40} = 15.708$ rad s$^{-1}$ — [M1] Keep the full value of $\omega$ for subsequent steps to avoid rounding errors.
2. $v_{max} = \omega A = 15.708 \times 0.050 = 0.785$ m s$^{-1}$ — [A1] Maximum speed occurs at $x = 0$. Rounding to 2 s.f.: $v_{max} \approx 0.79$ m s$^{-1}$.
3. $\omega^2 = 15.708 \times 15.708 = 246.7$ rad$^2$ s$^{-2}$ — [M1] Expanding $\omega^2$ longhand: $15.708 \times 15.708 \approx 246.7$. Do not round $\omega$ before squaring.
4. $a_{max} = \omega^2 A = 246.7 \times 0.050 = 12.3$ m s$^{-2}$ — [A1] Maximum acceleration occurs at $x = \pm A$. Rounding to 2 s.f.: $a_{max} \approx 12$ m s$^{-2}$.

Answer: (a) $v_{max} \approx 0.79$ m s$^{-1}$; (b) $a_{max} \approx 12$ m s$^{-2}$

6. Connecting Displacement, Velocity, and Acceleration Graphically

Earlier you saw the cosine shape of the displacement–time graph. The velocity–time graph follows from differentiating $x = A\cos(\omega t)$ with respect to time, giving $v = -A\omega\sin(\omega t)$. This means velocity is a negative sine wave — it is $90°$ (a quarter cycle) behind the displacement curve.

Differentiating once more gives acceleration: $a = -A\omega^2\cos(\omega t) = -\omega^2 x$. The acceleration–time graph is an inverted cosine — it is always in the opposite direction to displacement, which is precisely what the defining equation $a = -\omega^2 x$ requires.

Three key observations from this graph: when displacement is at its positive peak, velocity is zero and acceleration is at its negative peak. When displacement passes through zero going in the negative direction, velocity is at its negative peak and acceleration is zero. These phase relationships are a common source of exam questions.

7. Exam Technique and Common Mistakes

AQA SHM questions follow predictable patterns. The most reliable way to gain full marks is to follow a consistent method: identify what you are given, select the correct equation, show every substitution with units, and state the final answer to an appropriate number of significant figures.

One of the most common errors is forgetting to set the calculator to radians mode before evaluating $\cos(\omega t)$. The argument $\omega t$ is always in radians, never degrees. A quick sanity check: if $t = T/4$, then $\omega t = (2\pi/T)(T/4) = \pi/2$ radians, and $\cos(\pi/2) = 0$. If your calculator gives a non-zero answer for this, it is in degrees mode.

Every spec requirement in this lesson — the displacement–time equation, the velocity–displacement equation, maximum speed, and maximum acceleration — connects back to a single physical picture: an object oscillating symmetrically about an equilibrium position, trading between maximum displacement and maximum speed with every quarter cycle.

For SHM, the displacement-time equation is x = A cos(ωt) when the particle starts at maximum displacement. A useful linked result is v = ±ω√(A^2 − x^2), which shows that the speed depends on position: it is zero at the turning points where x = ±A, reaches a maximum value of ωA at the equilibrium position, and the maximum acceleration is ω^2A at the extremes.