SHM Equations

Now that we have established the defining equation of SHM (a = −ω²x), we need to develop the full set of equations that describe how displacement, velocity, and acceleration vary with time. These equations are the tools you will use to solve virtually every SHM problem at A-Level.

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Displacement as a Function of Time

The displacement of an object undergoing SHM varies sinusoidally with time. Depending on the starting conditions, we use either:

x = A cos(ωt) (if the object starts at maximum displacement)

x = A sin(ωt) (if the object starts at the equilibrium position moving in the positive direction)

where:

• x is the displacement at time t (m)
• A is the amplitude — the maximum displacement (m)
• ω is the angular frequency (rad s⁻¹)
• t is the time elapsed from the starting point (s)

The choice between cosine and sine depends on where the object is at t = 0. If you pull a mass on a spring to its maximum extension and release it, x starts at A, so the cosine form is natural. If you give the object a push through the equilibrium position, x starts at 0 and increases, so the sine form fits.

Both forms describe identical motion — they are simply shifted in time (phase) relative to each other. In fact, cos(ωt) = sin(ωt + π/2), so the cosine form leads the sine form by a quarter cycle.

Exam tip: Always check your calculator is in radians when evaluating SHM equations. Using degrees is one of the most common errors and will give completely wrong answers.

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Velocity as a Function of Time

The velocity is the rate of change of displacement with time. Differentiating the displacement equation:

If x = A cos(ωt), then:

v = dx/dt = −Aω sin(ωt)

If x = A sin(ωt), then:

v = dx/dt = Aω cos(ωt)

The maximum velocity occurs when the sine or cosine term equals ±1:

v_max = Aω

This maximum velocity occurs as the object passes through the equilibrium position (x = 0), where all the energy is kinetic. At the extreme positions (x = ±A), the velocity is zero momentarily as the object reverses direction.

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Acceleration as a Function of Time

Differentiating velocity gives acceleration:

If x = A cos(ωt), then:

a = dv/dt = −Aω² cos(ωt)

This can be rewritten as:

a = −ω²x

which is exactly the defining equation of SHM. The maximum acceleration occurs at maximum displacement:

a_max = Aω²

At equilibrium (x = 0), the acceleration is zero because there is no net restoring force.

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Velocity as a Function of Displacement

There is an extremely useful equation that relates velocity directly to displacement without involving time:

v = ±ω√(A² − x²)

This equation is derived from the energy relationships in SHM (which we cover in the next lesson), but it is worth learning now because it allows you to find the velocity at any displacement without needing to know the time.

Key features of this equation:

• When x = 0 (equilibrium): v = ±ωA (maximum speed)
• When x = ±A (maximum displacement): v = 0 (momentarily stationary)
• The ± sign indicates that the object can be moving in either direction at any given displacement (except at the extremes)

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Summary of Phase Relationships

The displacement, velocity, and acceleration are all sinusoidal functions of time, but they are out of phase with each other:

Quantity	When at maximum positive value	When zero	When at maximum negative value
Displacement x	At positive amplitude (+A)	At equilibrium	At negative amplitude (−A)
Velocity v	Passing through equilibrium (moving positive)	At ±A (momentarily stationary)	Passing through equilibrium (moving negative)
Acceleration a	At −A (pointing towards equilibrium)	At equilibrium	At +A (pointing towards equilibrium)

Velocity leads displacement by π/2 radians (a quarter cycle). Acceleration leads velocity by π/2 radians, meaning acceleration and displacement are π radians (half a cycle) out of phase — they are in antiphase. When displacement is maximum positive, acceleration is maximum negative, and vice versa.

flowchart TD
    A["Phase Relationships in SHM"] --> B["Displacement x"]
    A --> C["Velocity v"]
    A --> D["Acceleration a"]
    B -->|"x leads v by π/2?\nNo — v leads x by π/2"| C
    C -->|"v leads a by π/2?\nNo — a leads v by π/2"| D
    B -->|"x and a are in\nantiphase (π rad apart)"| D
    B --- E["At x = +A:\nv = 0, a = −ω²A"]
    C --- F["At x = 0:\nv = ±Aω, a = 0"]
    D --- G["At x = −A:\nv = 0, a = +ω²A"]

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Graphical Representations

If you plot x, v, and a against time on the same axes:

• The displacement curve is a cosine wave (assuming x = A cos(ωt))
• The velocity curve is a negative sine wave, reaching its peak a quarter period before the displacement reaches its peak
• The acceleration curve is a negative cosine wave — an inverted version of the displacement curve

These graphs are frequently asked about in exams. Make sure you can sketch all three on the same time axis and label the phase relationships correctly.

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Complete SHM Equations Reference

Starting condition	x	v	a
Starts at x = +A (released from max displacement)	A cos(ωt)	−Aω sin(ωt)	−Aω² cos(ωt)
Starts at x = 0 moving in +ve direction	A sin(ωt)	Aω cos(ωt)	−Aω² sin(ωt)

Quantity	Maximum value	Where it occurs
Displacement	A	At the extremes
Speed	Aω	At equilibrium (x = 0)
Acceleration magnitude	Aω²	At the extremes (x = ±A)

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Worked Example 1

A mass oscillates in SHM with amplitude 0.05 m and frequency 4 Hz. At t = 0 the mass is at maximum positive displacement.

Find: (a) the displacement at t = 0.1 s, (b) the maximum velocity, (c) the velocity at x = 0.03 m, (d) the maximum acceleration.

Solution:

ω = 2πf = 2π × 4 = 8π rad s⁻¹

(a) x = A cos(ωt) = 0.05 × cos(8π × 0.1) = 0.05 × cos(0.8π) = 0.05 × (−0.809) = −0.040 m

(b) v_max = Aω = 0.05 × 8π = 0.4π ≈ 1.26 m s⁻¹

(c) v = ±ω√(A² − x²) = ±8π × √(0.05² − 0.03²) = ±8π × √(0.0025 − 0.0009) = ±8π × √0.0016 = ±8π × 0.04 = ±0.32π ≈ ±1.01 m s⁻¹

(d) a_max = ω²A = (8π)² × 0.05 = 64π² × 0.05 ≈ 31.6 m s⁻²

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Worked Example 2

A particle undergoing SHM has a maximum acceleration of 12 m s⁻² and a maximum speed of 3 m s⁻¹. Find the amplitude, angular frequency, and period.

Solution:

We know a_max = ω²A = 12 and v_max = ωA = 3.

Dividing: a_max/v_max = ω²A/(ωA) = ω

So ω = 12/3 = 4 rad s⁻¹

Then A = v_max/ω = 3/4 = 0.75 m

T = 2π/ω = 2π/4 = π/2 ≈ 1.57 s

This technique of dividing a_max by v_max to find ω is extremely useful and appears frequently in exam questions.

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Worked Example 3

The velocity of an SHM oscillator is v = −0.8 sin(5t) m s⁻¹. Determine the amplitude, angular frequency, displacement equation, and the acceleration at t = π/10 s.

Solution:

Comparing v = −Aω sin(ωt) with v = −0.8 sin(5t):

ω = 5 rad s⁻¹ and Aω = 0.8, so A = 0.8/5 = 0.16 m

Since v = −Aω sin(ωt), the displacement must be x = A cos(ωt) = 0.16 cos(5t) m

At t = π/10: x = 0.16 cos(5 × π/10) = 0.16 cos(π/2) = 0.16 × 0 = 0 m

a = −ω²x = −25 × 0 = 0 m s⁻²

(The object is passing through equilibrium, so acceleration is zero and speed is at its maximum.)

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Worked Example 4: Finding Time to Reach a Given Displacement

A mass starts at x = +A = 0.06 m and oscillates with ω = 10 rad s⁻¹. How long does it take to first reach x = 0.03 m?

Solution:

x = A cos(ωt)

0.03 = 0.06 cos(10t)

cos(10t) = 0.5

10t = cos⁻¹(0.5) = π/3 rad

t = π/30 ≈ 0.105 s

Note: cos⁻¹ gives the first positive solution. There are multiple solutions (the mass returns to x = 0.03 m repeatedly), but this is the first time after t = 0.

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Common Exam Mistakes with SHM Equations

Mistake	Why it is wrong	How to avoid it
Calculator in degrees	SHM equations use radians	Always set calculator to RAD mode
Forgetting ± in v = ±ω√(A² − x²)	Object can move in either direction at a given x	Include ± unless the question specifies direction
Using sin when cos is needed (or vice versa)	Wrong initial conditions	Check: at t = 0, does x = A (cos) or x = 0 (sin)?
Writing v_max = Aω²	Confusing v_max with a_max	v_max = Aω (no squared); a_max = Aω²
Mixing up f and ω	f is in Hz, ω is in rad s⁻¹	Always check: ω = 2πf, not ω = f

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Real-World Application: Ultrasound Imaging

In medical ultrasound, a piezoelectric crystal oscillates in SHM at frequencies between 1 and 15 MHz. The crystal converts electrical oscillations to mechanical vibrations and vice versa. The SHM equations govern the displacement and velocity of the crystal surface, which determines the amplitude and frequency of the ultrasound wave. Understanding how to calculate v_max and a_max at such high frequencies explains why ultrasound can penetrate tissue: the tiny amplitude (nanometres) combined with very high ω produces the mechanical vibrations needed for imaging.

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A-Level Deep Dive: SHM Equations

Spec mapping

Edexcel 9PH0 specification Topic 13 — Oscillations requires candidates to use the kinematic equations $x = A\cos(\omega t)$x=Acos(ωt) (and the equivalent sine form), the velocity equation $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​, and the relationship $a = -\omega^2 x$a=−ω2x, together with the maxima $v_{\max} = A\omega$vmax​=Aω and $a_{\max} = A\omega^2$amax​=Aω2 (refer to the official specification document for exact wording). The equations are listed in the formula booklet, but the interpretation — choosing between sine and cosine according to initial conditions, recognising when to use the time-free form, and reasoning about phase relationships — is examined as application (AO2) rather than recall (AO1). Topic 13 sits in Paper 2, where SHM kinematics are routinely co-tested with circular motion (Topic 6, the projection link), with energy conservation (the $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​ equation has its origin in $\tfrac{1}{2}mv^2 + \tfrac{1}{2}m\omega^2 x^2 = \tfrac{1}{2}m\omega^2 A^2$21​mv2+21​mω2x2=21​mω2A2), and with graphical interpretation. Calculator mode (radians, never degrees) is an implicit requirement of every numerical SHM question.

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Worked example with full mark scheme

Question (8 marks):

A particle undergoes SHM with amplitude $A = 0.080\text{ m}$A=0.080 m and angular frequency $\omega = 5.0\text{ rad s}^{-1}$ω=5.0 rad s−1. At $t = 0$t=0 the particle is at $x = +A$x=+A.

(a) Write down the equations for $x(t)$x(t), $v(t)$v(t) and $a(t)$a(t). (3)

(b) Calculate the displacement, velocity and acceleration at $t = 0.20\text{ s}$t=0.20 s. (3)

(c) State the maximum speed and maximum acceleration, and the displacements at which each occurs. (2)

Solution with mark scheme:

(a) Because the particle starts at maximum positive displacement, the cosine form applies:

$x(t) = A\cos(\omega t) = 0.080\cos(5.0\,t)$x(t)=Acos(ωt)=0.080cos(5.0t)

Differentiating with respect to $t$t:

$v(t) = -A\omega\sin(\omega t) = -0.40\sin(5.0\,t)$v(t)=−Aωsin(ωt)=−0.40sin(5.0t) $a(t) = -A\omega^2\cos(\omega t) = -2.0\cos(5.0\,t)$a(t)=−Aω2cos(ωt)=−2.0cos(5.0t)

M1 — correct choice of cosine (initial condition recognised). A1 — correct $v(t)$v(t) with the leading minus sign and amplitude $A\omega = 0.40\text{ m s}^{-1}$Aω=0.40 m s−1. A1 — correct $a(t)$a(t) with amplitude $A\omega^2 = 2.0\text{ m s}^{-2}$Aω2=2.0 m s−2 and the minus sign retained.

(b) Step 1 — evaluate $\omega t$ωt in radians. $\omega t = 5.0 \times 0.20 = 1.0\text{ rad}$ωt=5.0×0.20=1.0 rad. (Calculator must be in radian mode.)

Step 2 — substitute.

$x = 0.080\cos(1.0) = 0.080 \times 0.5403 = 0.0432\text{ m}$x=0.080cos(1.0)=0.080×0.5403=0.0432 m $v = -0.40\sin(1.0) = -0.40 \times 0.8415 = -0.337\text{ m s}^{-1}$v=−0.40sin(1.0)=−0.40×0.8415=−0.337 m s−1 $a = -2.0\cos(1.0) = -2.0 \times 0.5403 = -1.08\text{ m s}^{-2}$a=−2.0cos(1.0)=−2.0×0.5403=−1.08 m s−2

M1 — correct numerical substitution. A1 — correct $x$x to 2 s.f. ($0.043\text{ m}$0.043 m). A1 — correct $v$v and $a$a with signs and units ($-0.34\text{ m s}^{-1}$−0.34 m s−1, $-1.1\text{ m s}^{-2}$−1.1 m s−2).

(c) Maximum speed $v_{\max} = A\omega = 0.080 \times 5.0 = 0.40\text{ m s}^{-1}$vmax​=Aω=0.080×5.0=0.40 m s−1, occurring at $x = 0$x=0 (equilibrium). Maximum acceleration magnitude $a_{\max} = A\omega^2 = 0.080 \times 25 = 2.0\text{ m s}^{-2}$amax​=Aω2=0.080×25=2.0 m s−2, occurring at $x = \pm A$x=±A (the extremes).

B1 — both maxima with units. B1 — both displacement locations correctly identified.

Total: 8 marks (M2 A4 B2 split as shown).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A mass on a spring oscillates in SHM with period $T = 0.50\text{ s}$T=0.50 s and amplitude $A = 0.040\text{ m}$A=0.040 m. At $t = 0$t=0 the mass passes through equilibrium moving in the positive direction.

(a) Calculate the angular frequency $\omega$ω and write the equation for $x(t)$x(t). (2)

(b) Find the speed of the mass when its displacement is $x = 0.020\text{ m}$x=0.020 m, using the time-free SHM equation. (3)

(c) State the next time after $t = 0$t=0 at which the mass is at maximum positive displacement. (1)

Mark scheme decomposition by AO:

(a)

• M1 (AO1) — $\omega = 2\pi/T = 2\pi/0.50 = 4\pi \approx 12.6\text{ rad s}^{-1}$ω=2π/T=2π/0.50=4π≈12.6 rad s−1.
• A1 (AO2) — initial condition (passing through equilibrium moving positively) implies sine form: $x(t) = 0.040\sin(4\pi t)$x(t)=0.040sin(4πt).

(b)

• M1 (AO1) — quoting the time-free equation $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​.
• M1 (AO2) — substituting: $v = \pm 4\pi\sqrt{0.040^2 - 0.020^2} = \pm 4\pi\sqrt{0.0012}$v=±4π0.0402−0.0202​=±4π0.0012​.
• A1 (AO1) — $|v| = 4\pi \times 0.0346 = 0.435\text{ m s}^{-1}$∣v∣=4π×0.0346=0.435 m s−1, i.e. $|v| \approx 0.44\text{ m s}^{-1}$∣v∣≈0.44 m s−1. The \pm\ sign should be retained or commented on (the mass passes through $x = 0.020\text{ m}$x=0.020 m in either direction).

(c)

• B1 (AO2) — starting at $x = 0$x=0 moving positively, the mass first reaches $+A$+A a quarter period later: $t = T/4 = 0.125\text{ s}$t=T/4=0.125 s.

Total: 6 marks split AO1 = 4, AO2 = 2. Edexcel reserves AO2 marks for the conceptual choices — sine versus cosine, identifying that the time-free equation bypasses the need for $t$t, and reading "next time at maximum positive displacement" as a quarter-period geometric fact rather than a calculation.

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