SHM Solutions — Displacement, Velocity and Acceleration

Spec mapping: OCR H556 Module 5.3 — Oscillations (sinusoidal solutions of $a = -\omega^2 x$a=−ω2x: $x = A\cos\omega t$x=Acosωt and $x = A\sin\omega t$x=Asinωt; velocity and acceleration as functions of time; $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​; phase and phase difference; $x$x-$t$t, $v$v-$t$t and $a$a-$t$t graph sketching). Refer to the official OCR H556 specification document for exact wording.

In the previous lesson we defined simple harmonic motion by the equation $a = -\omega^2 x$a=−ω2x. That is a differential equation in disguise: writing $a = d^2 x/dt^2$a=d2x/dt2, the defining equation becomes

$\frac{d^2 x}{dt^2} = -\omega^2 x$dt2d2x​=−ω2x

Solving this differential 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 must use confidently in OCR A-Level Physics A Module 5.3 — they generate every $x$x-$t$t, $v$v-$t$t, $a$a-$t$t graph, and every "where is the oscillator at $t = 0.3$t=0.3 s?" calculation that the exam will throw at you. The lesson concludes with the most useful equation in the whole SHM toolkit: $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​, which gives speed as a function of position (independent of time).

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The General Solution

Any function whose second derivative is equal to the original function multiplied by $-\omega^2$−ω2 must be a sine or cosine (or a linear combination). The two most useful forms for OCR are:

$x(t) = A\cos(\omega t) \qquad \text{— if the oscillator starts at } x = +A \text{ when } t = 0$x(t)=Acos(ωt)— if the oscillator starts at x=+A when t=0

$x(t) = A\sin(\omega t) \qquad \text{— if the oscillator starts at } x = 0, \text{ moving in the } +x \text{ direction, when } t = 0$x(t)=Asin(ωt)— if the oscillator starts at x=0, moving in the +x direction, when t=0

$A$A is the amplitude — the maximum displacement from equilibrium. $\omega$ω is the angular frequency we met in the last lesson, with units rad s$^{-1}$−1.

Both are solutions of $a = -\omega^2 x$a=−ω2x and both describe genuine SHM. Which one you use depends on where the oscillator was at $t = 0$t=0 — the initial condition. The OCR exam tends to favour the cosine form (initial condition: released from rest at the extreme), so that is what we will concentrate on here. A more general solution is $x(t) = A\cos(\omega t + \varphi)$x(t)=Acos(ωt+φ), where $\varphi$φ is the phase angle and lets you fix the initial condition arbitrarily.

Verification by substitution. Differentiate $x = A\cos(\omega t)$x=Acos(ωt) twice:

$\dot x = -A\omega\sin(\omega t), \qquad \ddot x = -A\omega^2\cos(\omega t) = -\omega^2 x$x˙=−Aωsin(ωt),x¨=−Aω2cos(ωt)=−ω2x

— which exactly satisfies the defining equation. The cosine and sine of $\omega t$ωt are the unique (up to amplitude and phase) functions whose second derivative reproduces the original function with a sign flip and a factor of $\omega^2$ω2. This is the deep mathematical reason for sinusoidal SHM solutions.

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Velocity of an SHM Oscillator

If $x = A\cos(\omega t)$x=Acos(ωt), then by differentiation

$v = \frac{dx}{dt} = -A\omega\sin(\omega t)$v=dtdx​=−Aωsin(ωt)

This tells us two things immediately:

1. The velocity is itself sinusoidal, with the same angular frequency $\omega$ω but shifted by a quarter cycle: the sine function is $\pi/2$π/2 ahead of cosine in argument-space.
2. The maximum speed is $|v_\text{max}| = A\omega$∣vmax​∣=Aω, occurring whenever $\sin(\omega t) = \pm 1$sin(ωt)=±1, i.e. when $\cos(\omega t) = 0$cos(ωt)=0, i.e. when the oscillator passes through equilibrium.

At the extremes of the motion, where $x = \pm A$x=±A, the velocity is zero — the oscillator momentarily stops before being pulled back. This is the universal pattern: fastest at the middle, stationary at the ends.

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Acceleration of an SHM Oscillator

Differentiating once more:

$a = \frac{dv}{dt} = -A\omega^2\cos(\omega t) = -\omega^2 x$a=dtdv​=−Aω2cos(ωt)=−ω2x

This is consistent with the defining equation, as it must be (it is the test of whether we have a correct solution). The maximum acceleration has magnitude

$|a_\text{max}| = A\omega^2$∣amax​∣=Aω2

and occurs at the extremes of the motion, exactly where the velocity is zero. At equilibrium ($x = 0$x=0) the acceleration is also zero — which makes physical sense, because the restoring force is zero at equilibrium.

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Summary of Maxima

Quantity	Formula	When does it occur?
Maximum displacement	$A$A	At the extremes of the motion
Maximum speed	$A\omega$Aω	Passing through equilibrium ($x = 0$x=0)
Maximum acceleration	$A\omega^2$Aω2	At the extremes of the motion ($x = \pm A$x=±A)

Memorising this table will save you time under exam conditions. Notice that each successive maximum picks up another factor of $\omega$ω — that is the signature of repeated differentiation of a sinusoid.

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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 $t$t between $x = A\cos(\omega t)$x=Acos(ωt) and $v = -A\omega\sin(\omega t)$v=−Aωsin(ωt) using the Pythagorean identity $\sin^2(\omega t) + \cos^2(\omega t) = 1$sin2(ωt)+cos2(ωt)=1:

$\sin^2(\omega t) = 1 - \cos^2(\omega t) = 1 - \left(\frac{x}{A}\right)^2$sin2(ωt)=1−cos2(ωt)=1−(Ax​)2

So

$v^2 = A^2\omega^2\sin^2(\omega t) = A^2\omega^2\left(1 - \frac{x^2}{A^2}\right) = \omega^2(A^2 - x^2)$v2=A2ω2sin2(ωt)=A2ω2(1−A2x2​)=ω2(A2−x2)

Taking the square root:

$v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​

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

• At $x = 0$x=0: $v = \pm\omega A$v=±ωA (maximum speed in either direction).
• At $x = \pm A$x=±A: $v = 0$v=0 (momentarily stationary at the turning points).
• At $x = A/2$x=A/2: $v = \pm\omega\sqrt{A^2 - A^2/4} = \pm\omega A\sqrt{3}/2 \approx 0.866\,\omega A$v=±ωA2−A2/4​=±ωA3​/2≈0.866ωA — that is, 86.6% of maximum speed at half the amplitude, not 50%.

This equation appears constantly in OCR exam papers and is the single most useful equation in the whole SHM toolkit. It is the equivalent of $v^2 = u^2 + 2as$v2=u2+2as for linear SUVAT — energy-conservation in disguise (lesson 5).

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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 $\varphi$φ and measured in radians.

• $x_1 = A\cos(\omega t)$x1​=Acos(ωt) and $x_2 = A\cos(\omega t)$x2​=Acos(ωt) are in phase — phase difference $0$0.
• $x_1 = A\cos(\omega t)$x1​=Acos(ωt) and $x_2 = A\cos(\omega t + \pi)$x2​=Acos(ωt+π) are in antiphase — phase difference $\pi$π (one is at $+A$+A when the other is at $-A$−A).
• $x_1 = A\cos(\omega t)$x1​=Acos(ωt) and $x_2 = A\sin(\omega t) = A\cos(\omega t - \pi/2)$x2​=Asin(ωt)=Acos(ωt−π/2) are $\pi/2$π/2 out of phase (a quarter cycle).

The displacement and velocity of a single oscillator are $\pi/2$π/2 out of phase (sine and cosine of the same argument), with $v$v leading $x$x by a quarter cycle. The displacement and acceleration are $\pi$π out of phase (in antiphase) — which is exactly the minus sign in $a = -\omega^2 x$a=−ω2x.

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The SHM Cycle — a Visual Summary

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

graph LR
    S0["t = 0<br/>x = +A<br/>v = 0<br/>a = −Aω²"] --> S1
    S1["t = T/4<br/>x = 0<br/>v = −Aω<br/>a = 0"] --> S2
    S2["t = T/2<br/>x = −A<br/>v = 0<br/>a = +Aω²"] --> S3
    S3["t = 3T/4<br/>x = 0<br/>v = +Aω<br/>a = 0"] --> S4
    S4["t = T<br/>x = +A<br/>v = 0<br/>a = −Aω²"]

Note the pattern:

• Displacement and acceleration are always equal in size and opposite in sign (the defining equation $a = -\omega^2 x$a=−ω2x).
• Velocity is maximal where displacement is zero, and zero where displacement is maximal.
• Each row in the table differs by $T/4$T/4 (a quarter period). The full cycle takes exactly $T$T.

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Worked Example 1 — Displacement at a Given Time

A pendulum swings with amplitude $0.12$0.12 m and period $2.0$2.0 s. Taking $x = +A$x=+A at $t = 0$t=0, find the displacement at $t = 0.30$t=0.30 s.

$\begin{aligned} \omega &= \frac{2\pi}{T} = \frac{2\pi}{2.0} = \pi \text{ rad s}^{-1} \\ x(0.30) &= 0.12\cos(\pi \times 0.30) = 0.12\cos(0.942) = 0.12 \times 0.588 = 0.0705 \text{ m} \end{aligned}$ωx(0.30)​=T2π​=2.02π​=π rad s−1=0.12cos(π×0.30)=0.12cos(0.942)=0.12×0.588=0.0705 m​

After $0.30$0.30 s the bob has moved from $+0.12$+0.12 m to about $+0.07$+0.07 m. That is $15\%$15% of a full period past the start, so we expect to be partway through the descent toward equilibrium — and $+0.07$+0.07 m looks right.

Make sure the calculator is in radian mode: $\cos(0.942\text{ rad}) = 0.588$cos(0.942 rad)=0.588, but $\cos(0.942°) = 0.99987$cos(0.942°)=0.99987 — a totally different number, which would give the wrong displacement by a factor of nearly $2$2.

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Worked Example 2 — Speed at a Given Displacement

A mass on a spring oscillates with amplitude $0.080$0.080 m and angular frequency $15$15 rad s$^{-1}$−1. Find the speed when (a) it passes through equilibrium and (b) it is $0.040$0.040 m from equilibrium.

(a) Maximum speed occurs at $x = 0$x=0:

$v_\text{max} = A\omega = 0.080 \times 15 = 1.20 \text{ m s}^{-1}$vmax​=Aω=0.080×15=1.20 m s−1

(b) Using $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​:

$\begin{aligned} v &= \pm 15\sqrt{0.080^2 - 0.040^2} \\ &= \pm 15\sqrt{0.00640 - 0.00160} \\ &= \pm 15\sqrt{0.00480} \\ &= \pm 15 \times 0.0693 \\ &= \pm 1.04 \text{ m s}^{-1} \end{aligned}$v​=±150.0802−0.0402​=±150.00640−0.00160​=±150.00480​=±15×0.0693=±1.04 m s−1​

At half the amplitude, the speed is about $87\%$87% of the maximum — not half. This is because $v^2 \propto (A^2 - x^2)$v2∝(A2−x2), not $(A - x)$(A−x). A useful general result: at $|x| = A/n$∣x∣=A/n, the speed is $v = v_\text{max}\sqrt{1 - 1/n^2}$v=vmax​1−1/n2​, which gives $0.866$0.866 at $n = 2$n=2 and $0.745$0.745 at $n = 1.5$n=1.5.

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Worked Example 3 — A Tuning Fork

A tuning fork vibrates at $256$256 Hz (musical middle C) with amplitude $0.12$0.12 mm. Calculate (a) the angular frequency, (b) the maximum speed of a prong, and (c) the maximum acceleration of a prong.

(a)

$\omega = 2\pi f = 2\pi \times 256 = 1\,608 \text{ rad s}^{-1}$ω=2πf=2π×256=1608 rad s−1

(b)

$v_\text{max} = A\omega = 1.2 \times 10^{-4} \times 1\,608 = 0.193 \text{ m s}^{-1}$vmax​=Aω=1.2×10−4×1608=0.193 m s−1

(c)

$|a_\text{max}| = A\omega^2 = 1.2 \times 10^{-4} \times (1\,608)^2 = 1.2 \times 10^{-4} \times 2.585 \times 10^6 = 310 \text{ m s}^{-2}$∣amax​∣=Aω2=1.2×10−4×(1608)2=1.2×10−4×2.585×106=310 m s−2

That is about $32\,g$32g — a useful reminder that very small amplitude motions at high frequencies still involve huge accelerations. The prongs of a tuning fork experience accelerations large enough to launch a marble across the room, in principle; their tiny mass means the forces on the metal itself remain modest, but the kinematics of the vibrating surface are extreme.

This is a general principle of high-frequency SHM: amplitude can be sub-millimetre, but $\omega^2 A$ω2A scales as $f^2$f2, and a $1$1 kHz vibration with $1$1 mm amplitude already involves $\sim 4000\,g$∼4000g accelerations.

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Worked Example 4 — Time to Reach a Position

A body in SHM has amplitude $0.25$0.25 m and period $4.0$4.0 s. Starting from $x = +A$x=+A at $t = 0$t=0, how long does it take to reach $x = +0.125$x=+0.125 m for the first time?

$\begin{aligned} \omega &= \frac{2\pi}{T} = \frac{2\pi}{4.0} = 1.571 \text{ rad s}^{-1} \\ x &= A\cos(\omega t) \\ 0.125 &= 0.25\cos(1.571\,t) \\ \cos(1.571\,t) &= 0.5 \\ 1.571\,t &= \frac{\pi}{3} \approx 1.047 \\ t &= \frac{1.047}{1.571} = 0.667 \text{ s} \end{aligned}$ωx0.125cos(1.571t)1.571tt​=T2π​=4.02π​=1.571 rad s−1=Acos(ωt)=0.25cos(1.571t)=0.5=3π​≈1.047=1.5711.047​=0.667 s​

A bit under a sixth of the way through a cycle ($T/6 = 0.667$T/6=0.667 s exactly). Sense-check: at half amplitude the oscillator is still going fast ($87\%$87% of $v_\text{max}$vmax​, from Example 2), so it has not spent long getting there from the extreme.

This is a useful general result: the time to drop from $x = A$x=A to $x = A/2$x=A/2 is exactly $T/6$T/6 for any SHM oscillator. The time to drop from $A$A to $0$0 is $T/4$T/4, and from $A$A to $-A$−A is $T/2$T/2. These three landmarks ($T/6$T/6, $T/4$T/4, $T/2$T/2) appear in many OCR questions.

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Worked Example 5 — Starting from Equilibrium

A mass on a spring is set in motion by being struck at equilibrium with speed $0.50$0.50 m s$^{-1}$−1, in the $+x$+x direction. The system has $\omega = 4.0$ω=4.0 rad s$^{-1}$−1. Find (a) the amplitude, (b) the position at $t = 0.50$t=0.50 s, and (c) the speed at $t = 0.50$t=0.50 s.

This is a sine-form initial condition: $x(0) = 0$x(0)=0 and $\dot x(0) > 0$x˙(0)>0. So write $x(t) = A\sin(\omega t)$x(t)=Asin(ωt).

(a) Differentiating, $v(t) = A\omega\cos(\omega t)$v(t)=Aωcos(ωt). At $t = 0$t=0: $v(0) = A\omega = 0.50$v(0)=Aω=0.50, so $A = 0.50/4.0 = 0.125$A=0.50/4.0=0.125 m.

(b) $x(0.50) = 0.125\sin(4.0 \times 0.50) = 0.125\sin(2.0) = 0.125 \times 0.909 = 0.114$x(0.50)=0.125sin(4.0×0.50)=0.125sin(2.0)=0.125×0.909=0.114 m.

(c) $v(0.50) = A\omega\cos(\omega t) = 0.50\cos(2.0) = 0.50 \times (-0.416) = -0.208$v(0.50)=Aωcos(ωt)=0.50cos(2.0)=0.50×(−0.416)=−0.208 m s$^{-1}$−1.

So at $t = 0.5$t=0.5 s the mass is at $x = +0.114$x=+0.114 m (close to the positive extreme of $0.125$0.125 m) and moving backward (negative speed of $0.21$0.21 m s$^{-1}$−1). This is consistent with the geometry: it has gone past the extreme of the swing and is on its way back through equilibrium.

Sense-check using $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​: $v = \pm 4.0\sqrt{0.125^2 - 0.114^2} = \pm 4.0\sqrt{0.0025} = \pm 0.20$v=±4.00.1252−0.1142​=±4.00.0025​=±0.20 m s$^{-1}$−1. Excellent match (the small discrepancy is rounding).

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Graphs of $x$x, $v$v and $a$a against Time

The three curves are sinusoids, all with the same angular frequency $\omega$ω, but shifted in phase by quarter cycles: