Simple Harmonic Motion

A-Level Physics · Further Mechanics and Thermal Physics

Simple Harmonic Motion

Simple harmonic motion (SHM) is a special type of oscillatory motion in which the acceleration is always directed toward a fixed equilibrium position and is proportional to the displacement from that position.

Defining Equation

a = −ω²x

where:

  • a = acceleration (m s⁻²)
  • ω = angular frequency (rad s⁻¹)
  • x = displacement from equilibrium (m)
  • The negative sign indicates the acceleration is always directed opposite to the displacement (toward equilibrium)

This is the defining condition for SHM. If a system satisfies this equation, it undergoes SHM.

Key Equations of SHM

Displacement: x = A cos(ωt) or x = A sin(ωt)

(The choice depends on initial conditions: cos if timing starts at maximum displacement, sin if starting at equilibrium.)

Velocity: v = −Aω sin(ωt) or v = Aω cos(ωt)

Maximum velocity: v_max = Aω (at equilibrium, x = 0)

Velocity as a function of displacement: v = ±ω√(A² − x²)

Acceleration: a = −Aω² cos(ωt) = −ω²x

Maximum acceleration: a_max = Aω² (at maximum displacement, x = ±A)

Period and Frequency

ω = 2πf = 2π/T

The period T is independent of amplitude — this is a key characteristic of SHM called isochronous oscillation.

Energy in SHM

Kinetic energy: E_k = ½mv² = ½mω²(A² − x²)

Potential energy: E_p = ½mω²x² (= ½kx² for a spring)

Total energy: E_total = ½mω²A² = constant

At equilibrium (x = 0): all energy is kinetic (E_k = max, E_p = 0)

At maximum displacement (x = ±A): all energy is potential (E_k = 0, E_p = max)

Energy is continuously exchanged between kinetic and potential forms, but the total mechanical energy remains constant (in the absence of damping).

The Mass-Spring System

A mass m on a spring of spring constant k:

Restoring force: F = −kx → ma = −kx → a = −(k/m)x

Comparing with a = −ω²x: ω² = k/m

Period: T = 2π√(m/k)

Derivation

From Newton's second law: m(d²x/dt²) = −kx

This gives d²x/dt² = −(k/m)x, which has the solution x = A cos(ωt) with ω = √(k/m).

Therefore T = 2π/ω = 2π√(m/k)

The Simple Pendulum

For a pendulum of length L making small angle oscillations (θ < ~10°):

The restoring force along the arc is F = −mg sin θ ≈ −mgθ (small angle approximation)

Since x = Lθ: F = −mg(x/L) → a = −(g/L)x

So ω² = g/L and T = 2π√(L/g)

Note: The period depends only on L and g, not on the mass or the amplitude (for small oscillations).

Worked Example

A 0.40 kg mass oscillates on a spring (k = 25 N m⁻¹) with amplitude 0.060 m. Find:

(a) Period:

T = 2π√(m/k) = 2π√(0.40/25) = 2π√(0.016) = 2π × 0.1265 = 0.795 s

(b) Maximum speed:

ω = 2π/T = 2π/0.795 = 7.90 rad s⁻¹

v_max = Aω = 0.060 × 7.90 = 0.474 m s⁻¹

(c) Maximum acceleration:

a_max = Aω² = 0.060 × 7.90² = 0.060 × 62.4 = 3.75 m s⁻²

(d) Speed when x = 0.030 m:

v = ω√(A² − x²) = 7.90 × √(0.060² − 0.030²) = 7.90 × √(0.0027) = 7.90 × 0.05196 = 0.410 m s⁻¹

Graphical Representation

The displacement, velocity, and acceleration graphs are all sinusoidal:

  • Velocity leads displacement by π/2 (90°) — v is maximum when x = 0
  • Acceleration leads velocity by π/2 and is in antiphase with displacement
  • The a-x graph is a straight line through the origin with gradient −ω² (this is the test for SHM)

Damped Oscillations

In real systems, energy is lost to the surroundings (friction, air resistance), and the amplitude decreases over time.

  • Light damping: Amplitude decreases exponentially; the system oscillates many times before stopping; period approximately unchanged
  • Heavy damping: Few oscillations before stopping; period slightly longer
  • Critical damping: Returns to equilibrium in the minimum time without oscillating (e.g., car shock absorbers)
  • Overdamping: Returns to equilibrium without oscillating but takes longer than critical damping (e.g., fire doors)

Forced Oscillations and Resonance

When a periodic driving force acts on an oscillating system:

  • If driving frequency << natural frequency: amplitude is small, oscillation in phase with driver
  • If driving frequency = natural frequency (f₀): resonance — maximum amplitude, phase lag = π/2
  • If driving frequency >> natural frequency: amplitude is small, oscillation in antiphase with driver

Required Practical: Investigating SHM of a Mass-Spring System

1. Attach masses to a spring and measure T for ~10 oscillations

2. Plot T² against m — should be a straight line through the origin

3. Gradient = 4π²/k → determine k

4. Verify T is independent of amplitude by varying A and measuring T

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