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Simple harmonic motion (SHM) is the linearised model of every oscillation in nature. A mass on a spring, a pendulum at small angle, the vibrating diatomic molecule, the electron in an atom under a restoring electrostatic force, the LC tuning circuit in a radio receiver, every musical note from every instrument — all are SHM to first approximation, and the same single defining equation describes them. This lesson states that equation, derives the position, velocity and acceleration as functions of time, and links the kinematic descriptions to circular motion through the reference-circle construction.
Spec mapping: This lesson sits under AQA 7408 section 3.6.1.2. It covers the defining condition for SHM (a = −ω²x), the displacement-time, velocity-time and acceleration-time descriptions x = A cos(ωt + φ), v = −Aω sin(ωt + φ), a = −Aω² cos(ωt + φ), the period T = 2π/ω, the relationship v = ±ω√(A² − x²), and the isochronous nature of SHM (period independent of amplitude). (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.6.1.1 (circular motion): SHM is the projection of uniform circular motion onto a diameter. The angular frequency ω in SHM is literally the angular speed of the reference circle.
- Section 3.3.1 (progressive waves): every point on a transverse wave executes SHM in the transverse direction. The angular frequency ω of a wave's wave equation y(x,t) = A sin(kx − ωt + φ) is the SHM angular frequency at any fixed x.
- Section 3.5 (electricity — capacitors and inductors): the LC oscillator discharges and recharges according to d²Q/dt² = −Q/(LC), which is mathematically identical to SHM with ω² = 1/(LC). The resonance phenomena in radio tuning circuits are SHM at a different physical level.
Many oscillatory systems satisfy a common dynamical pattern: when displaced from equilibrium, they experience a restoring force that pulls them back, and the strength of that force is proportional to how far they have been displaced. In symbols, the restoring force on a body of mass m at displacement x from equilibrium is
F_restoring = −kx (for some positive constant k)
where the minus sign signals that the force is directed opposite to the displacement (i.e. towards equilibrium). By Newton's second law F = ma:
ma = −kx → a = −(k/m)x.
Defining ω² = k/m (with ω being the angular frequency of the oscillation, units rad s⁻¹) gives the defining equation of SHM:
a = −ω²x
This is the single most important equation in oscillation theory. It says:
Definition. A body undergoes simple harmonic motion when its acceleration is proportional to its displacement from a fixed equilibrium position and directed always towards that equilibrium position: a = −ω²x.
To test whether a given system performs SHM, derive its acceleration as a function of displacement. If the relationship is linear with a negative coefficient, the motion is SHM, and the magnitude of the coefficient is ω². If the relationship is non-linear (a quadratic restoring force, a cubic, a step), the motion is not SHM — it might be approximately SHM for small displacements (the linearised regime), but exact SHM requires linearity.
The equation a = −ω²x is a second-order linear ordinary differential equation in time, since a = d²x/dt². The general solution can be written in any of three equivalent forms:
| Form | Expression |
|---|---|
| Cosine with phase | x(t) = A cos(ωt + φ) |
| Sine with phase | x(t) = A sin(ωt + ψ) |
| Cosine + sine | x(t) = C cos(ωt) + D sin(ωt) |
These three forms are equivalent — any one can be transformed into another by appropriate choice of amplitude and phase. AQA convention is to use x = A cos(ωt + φ) as the standard solution. We will adopt that convention here.
x(t) = A cos(ωt + φ)
At t = 0: x(0) = A cos φ. If φ = 0, the body starts at the positive extreme x = +A. If φ = π/2, x(0) = 0 (the body is passing through equilibrium). The phase constant simply shifts the cosine left or right in time.
Differentiating x = A cos(ωt + φ) with respect to time:
v(t) = dx/dt = −Aω sin(ωt + φ)
The maximum magnitude of velocity occurs when sin(ωt + φ) = ±1, i.e. at the equilibrium position (where x = 0). The peak speed is
v_max = Aω
and it occurs each time the body passes through equilibrium. At the extremes (x = ±A), the speed is zero — the body momentarily reverses direction.
Differentiating v = −Aω sin(ωt + φ) with respect to time:
a(t) = dv/dt = −Aω² cos(ωt + φ) = −ω²x
The acceleration is largest when |x| is largest, i.e. at the extremes. The peak acceleration is
a_max = Aω²
and it occurs at x = ±A. At equilibrium (x = 0) the acceleration is zero — the body is moving at peak speed but is, for that instant, neither speeding up nor slowing down.
The textbook identity
v = ±ω√(A² − x²)
links the speed at any displacement x to the amplitude A and the angular frequency ω. It is derived by eliminating t from the cosine and sine expressions: x/A = cos(ωt + φ) and v/(Aω) = −sin(ωt + φ). Squaring and adding gives (x/A)² + (v/Aω)² = 1, which rearranges to v² = ω²(A² − x²).
This form is useful in exam questions where time is not given. Typical applications:
The plus/minus sign reflects that the body passes through each displacement x twice per period — once moving forward and once moving backward — at the same speed.
A body undergoes SHM with amplitude 0.080 m and angular frequency 25 rad s⁻¹. Calculate (a) the maximum speed, (b) the speed at displacement 0.060 m, (c) the magnitude of the acceleration at this displacement.
(a) v_max = ωA = 25 × 0.080 = 2.0 m s⁻¹.
(b) v = ω√(A² − x²) = 25 × √(0.080² − 0.060²) = 25 × √(0.0064 − 0.0036) = 25 × √0.0028 = 25 × 0.0529 = 1.32 m s⁻¹.
(c) |a| = ω²x = 25² × 0.060 = 625 × 0.060 = 37.5 m s⁻².
Compare the SHM solution
x(t) = A cos(ωt + φ)
with the x-coordinate of a point moving in uniform circular motion of radius A at angular speed ω, starting at angle φ:
x_circle(t) = A cos(ωt + φ).
They are identical. This is the reference-circle construction: SHM is precisely the projection of uniform circular motion onto a diameter. Imagine a point P moving steadily around a circle of radius A in the x–y plane; its shadow on the x-axis (the projection of its position onto the x-axis) performs exact SHM with amplitude A and angular frequency ω.
This construction is more than a pretty analogy. It explains:
Every kinematic formula of circular motion has an SHM counterpart, and most of the algebra is identical with v_circle = ωA playing the role of v_max and a_circle = ω²A playing the role of a_max.
graph LR
A["Reference circle<br/>radius A, angular speed ω"] --> B["Point P at angle ωt + φ"]
B --> C["x-coordinate of P:<br/>x = A cos(ωt + φ)"]
C --> D["SHM oscillator on x-axis:<br/>x = A cos(ωt + φ)"]
B --> E["Tangential v_circle = Aω"]
E --> F["v_max of SHM = Aω<br/>(at equilibrium)"]
B --> G["Centripetal a_circle = Aω²"]
G --> H["a_max of SHM = Aω²<br/>(at extremes)"]
style D fill:#27ae60,color:#fff
style F fill:#3498db,color:#fff
style H fill:#e67e22,color:#fff
A striking feature of SHM is that the period is independent of the amplitude:
T = 2π/ω = 2π√(m/k) (for a mass-spring system)
The amplitude A does not appear in the formula for T. Whether you displace a mass-spring system by 1 cm or 5 cm, the period of oscillation is the same. This property is called isochronism ("equal-time" in Greek), and it is responsible for the use of pendulum clocks (since at small amplitudes a pendulum is approximately SHM and so approximately isochronous). The historical attribution to Galileo's observation of the chandelier in Pisa Cathedral is part of physics folklore; what matters for the exam is the property itself and its underlying cause — the linear restoring force.
Isochronism fails for non-SHM oscillators. A pendulum at large amplitude, where sin θ ≠ θ, has a period that grows slightly with amplitude (about 0.4% slower at 30° than at 5°). The Foucault pendulum at the Panthéon in Paris is intentionally operated at small amplitude to preserve isochronism for precise period measurement.
A spring with k = 80 N m⁻¹ supports a mass of 0.50 kg. Calculate the period (a) for a 2.0 cm initial displacement and (b) for a 10.0 cm initial displacement.
ω² = k/m = 80/0.50 = 160 → ω = 12.65 rad s⁻¹.
(a) T = 2π/ω = 2π/12.65 = 0.497 s at 2 cm amplitude.
(b) T = 2π/ω = 2π/12.65 = 0.497 s at 10 cm amplitude.
The period is the same in both cases. Only the maximum speed (v_max = ωA, scaling linearly with A) and the maximum acceleration (a_max = ω²A) differ.
The canonical SHM oscillator is a mass on a horizontal spring. Let a block of mass m be attached to a spring of spring constant k, with the equilibrium position at x = 0. When the block is displaced to x, Hooke's law gives the restoring force as F = −kx, so by Newton's second law:
ma = −kx → a = −(k/m)x.
Comparing with the defining equation a = −ω²x:
ω² = k/m → ω = √(k/m) → T = 2π/ω = 2π√(m/k).
This is the standard period formula for a mass-spring system. Heavier masses oscillate more slowly (T ∝ √m); stiffer springs oscillate more quickly (T ∝ 1/√k). The amplitude A drops out — confirming isochronism.
Lesson 5 develops the mass-spring system and the simple pendulum in full. For now, recognise that the simple structure "F = −kx → a = −ω²x → T = 2π√(m/k)" is the template for analysing every SHM oscillator: identify the linear restoring force, equate F = ma, and read off ω².
An object oscillates with SHM such that x(t) = 0.05 cos(4π t) m, where t is in seconds. (a) Identify the amplitude, angular frequency, period, and frequency. (b) Calculate the maximum velocity and maximum acceleration. (c) Calculate x, v, and a at t = 0.10 s, and sense-check the sign and magnitude of each.
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