Simple Harmonic Motion

a = -ω²x; x = A cos(ωt); energy in SHM; pendulums and mass-spring systems

# Simple Harmonic Motion — A-Level Physics

Simple Harmonic Motion (SHM) describes oscillating systems — from pendulums to atoms vibrating in a crystal. It's one of the most important and widely applicable concepts in physics.


1. Definition of SHM

SHM occurs when:

  • The acceleration is always directed towards a fixed equilibrium point
  • The acceleration is proportional to the displacement from equilibrium

a=ω2x\boxed{a = -\omega^2 x}

The negative sign means the acceleration is always in the opposite direction to the displacement (restoring force).


2. Key Equations

Displacement

x=Acos(ωt)orx=Asin(ωt)x = A\cos(\omega t) \quad \text{or} \quad x = A\sin(\omega t)

(Use cosine if timing starts from maximum displacement; sine if from equilibrium.)

Velocity

v=Aωsin(ωt)orv=Aωcos(ωt)v = -A\omega\sin(\omega t) \quad \text{or} \quad v = A\omega\cos(\omega t)

Maximum velocity: vmax=Aωv_{\text{max}} = A\omega (at equilibrium, x=0x = 0)

Alternative form: v=±ωA2x2v = \pm\omega\sqrt{A^2 - x^2}

Acceleration

a=Aω2cos(ωt)=ω2xa = -A\omega^2\cos(\omega t) = -\omega^2 x

Maximum acceleration: amax=Aω2a_{\text{max}} = A\omega^2 (at maximum displacement, x=±Ax = \pm A)


3. Period and Frequency

ω=2πf=2πT\omega = 2\pi f = \frac{2\pi}{T}

Mass-Spring System

T=2πmk\boxed{T = 2\pi\sqrt{\frac{m}{k}}}

ω=k/m\omega = \sqrt{k/m}

  • Period depends on mass and spring constant
  • Period does NOT depend on amplitude

Simple Pendulum

T=2πLg\boxed{T = 2\pi\sqrt{\frac{L}{g}}}

ω=g/L\omega = \sqrt{g/L}

  • Period depends on length and gg
  • Period does NOT depend on mass or amplitude (for small angles, θ<10°\theta < 10°)

4. Energy in SHM

KE=12mω2(A2x2)KE = \frac{1}{2}m\omega^2(A^2 - x^2) PE=12mω2x2PE = \frac{1}{2}m\omega^2 x^2 Etotal=12mω2A2=constantE_{\text{total}} = \frac{1}{2}m\omega^2 A^2 = \text{constant}

Key Relationships

  • At x=0x = 0 (equilibrium): KE is maximum, PE is minimum (zero)
  • At x=±Ax = \pm A (extremes): KE is zero, PE is maximum
  • Total energy is constant (for undamped SHM)
  • KE and PE oscillate between 0 and EtotalE_{\text{total}}
  • Energy graphs show KE and PE as sinusoidal, with KE + PE = constant

5. Graphs of SHM

Graph Shape Key Features
xx vs tt Cosine/sine Amplitude AA, period TT
vv vs tt Sine/cosine (90° ahead of xx) Max at equilibrium
aa vs tt Cosine/sine (180° ahead of xx) Max at extremes
aa vs xx Straight line through origin Gradient = ω2-\omega^2
vv vs xx Ellipse Max vv at x=0x = 0

6. Damping

Damped oscillations lose energy over time (to friction, air resistance).

Type Description
Light damping Amplitude decreases gradually; frequency barely changes
Heavy damping Returns to equilibrium slowly without oscillating
Critical damping Returns to equilibrium as quickly as possible without oscillating

Critical damping is used in car shock absorbers and galvanometers.


7. Forced Oscillations and Resonance

Forced oscillation: An external periodic force drives the system.

Resonance occurs when the driving frequency equals the natural frequency of the system. At resonance:

  • Amplitude is maximum
  • Energy transfer from driver to oscillator is most efficient

Effects of damping on resonance:

  • Lower peak amplitude
  • Broader resonance peak
  • Resonant frequency shifts slightly lower

Examples of resonance:

  • Tuning a radio (electrical resonance)
  • Tacoma Narrows Bridge collapse (wind-driven resonance)
  • Microwave ovens (resonance of water molecules)
  • Musical instruments (air column/string resonance)

Worked Example: Mass-Spring

Problem

A 0.5 kg mass on a spring (k=200k = 200 N/m) oscillates with amplitude 3 cm. Find T, vmaxv_{\text{max}}, amaxa_{\text{max}}.

T=2πm/k=2π0.5/200=2π(0.05)=0.314T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/200} = 2\pi(0.05) = 0.314 s ω=2π/T=20\omega = 2\pi/T = 20 rad/s vmax=Aω=0.03×20=0.6v_{\text{max}} = A\omega = 0.03 \times 20 = 0.6 m/s amax=Aω2=0.03×400=12a_{\text{max}} = A\omega^2 = 0.03 \times 400 = 12 m/s²

Solution

Worked Example: Velocity at a Point

Problem

In the above system, find the velocity when x=2x = 2 cm.

v=ωA2x2=200.0320.022=200.0005=20×0.0224=0.447v = \omega\sqrt{A^2 - x^2} = 20\sqrt{0.03^2 - 0.02^2} = 20\sqrt{0.0005} = 20 \times 0.0224 = 0.447 m/s

Solution

Worked Example: Simple Pendulum

Problem

A pendulum has a period of 1.5 s. Find its length.

T=2πL/gT = 2\pi\sqrt{L/g}L=gT2/(4π2)=9.81×2.25/(4π2)=22.07/39.48=0.559L = gT^2/(4\pi^2) = 9.81 \times 2.25/(4\pi^2) = 22.07/39.48 = 0.559 m


Solution

9. Practice Questions

    1. Show that a mass on a spring undergoes SHM by deriving a=ω2xa = -\omega^2 x. (3 marks)
    1. A particle oscillates with SHM of amplitude 5 cm and period 0.4 s. Calculate (a) ω\omega, (b) max velocity, (c) max acceleration. (4 marks)
    1. Sketch graphs of displacement, velocity, and acceleration against time for SHM, and state the phase relationships. (3 marks)
    1. Explain what happens when the driving frequency of a forced oscillation approaches the natural frequency. How does damping affect this? (4 marks)

    Answers

    1. Spring force: F=kxF = -kx. By N2: ma=kxma = -kxa=(k/m)xa = -(k/m)x. Since k/m=ω2k/m = \omega^2: a=ω2xa = -\omega^2 x. ✓ SHM.

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Summary

  • SHM: a=ω2xa = -\omega^2 x (acceleration proportional to displacement, towards equilibrium)
  • x=Acos(ωt)x = A\cos(\omega t), v=±ωA2x2v = \pm\omega\sqrt{A^2 - x^2}, vmax=Aωv_{\max} = A\omega, amax=Aω2a_{\max} = A\omega^2
  • Mass-spring: T=2πm/kT = 2\pi\sqrt{m/k}; Pendulum: T=2πL/gT = 2\pi\sqrt{L/g}
  • Energy: Etotal=12mω2A2E_{\text{total}} = \frac{1}{2}m\omega^2A^2 = constant (undamped)
  • Resonance: driving frequency = natural frequency → max amplitude
  • Damping reduces amplitude; critical damping = fastest return without oscillation

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