Oscillations and Wave Characteristics

SHM; wave speed v = fλ; transverse vs longitudinal; amplitude, period, frequency

# Oscillations and Wave Characteristics — IB Physics

Waves and oscillations are everywhere — from sound to light, from pendulums to atoms vibrating in crystals. Understanding wave properties is essential for many areas of IB Physics.


1. Simple Harmonic Motion (SHM)

An oscillation is SHM if the restoring force (and acceleration) is proportional to displacement and directed towards the equilibrium position:

a=ω2xa = -\omega^2 x

x=Acos(ωt)x = A\cos(\omega t); v=Aωsin(ωt)v = -A\omega\sin(\omega t); a=Aω2cos(ωt)a = -A\omega^2\cos(\omega t)

Where ω=2πf=2π/T\omega = 2\pi f = 2\pi/T.

vmax=Aωv_{\max} = A\omega (at x=0x = 0); amax=Aω2a_{\max} = A\omega^2 (at x=±Ax = \pm A).

Energy in SHM

ET=12mω2A2E_T = \frac{1}{2}m\omega^2 A^2 (constant for undamped SHM) At x=0x = 0: all KE. At x=±Ax = \pm A: all PE.

2. Wave Properties

Property Definition
Amplitude (AA) Max displacement from equilibrium
Wavelength (λ\lambda) Distance between consecutive points in phase
Frequency (ff) Oscillations per second (Hz)
Period (TT) Time for one complete oscillation; T=1/fT = 1/f
Wave speed v=fλv = f\lambda

3. Transverse vs Longitudinal

Transverse: Oscillation perpendicular to direction of propagation (EM waves, water waves, string waves). Can be polarised.

Longitudinal: Oscillation parallel to propagation (sound, P-waves). Compressions and rarefactions. Cannot be polarised.

4. Wave Equation

y=Asin(ωtkx)y = A\sin(\omega t - kx)

Where k=2π/λk = 2\pi/\lambda (wave number), ω=2πf\omega = 2\pi f.

5. Intensity

I=PAA2I = \frac{P}{A} \propto A^2

For a point source: I1/r2I \propto 1/r^2 (inverse square law).

Worked Example: SHM

Problem

A mass on a spring oscillates with A=4A = 4 cm and f=2f = 2 Hz. Find vmaxv_{\max} and amaxa_{\max}.

ω=2π(2)=12.57\omega = 2\pi(2) = 12.57 rad/s vmax=Aω=0.04(12.57)=0.503v_{\max} = A\omega = 0.04(12.57) = 0.503 m/s amax=Aω2=0.04(158)=6.32a_{\max} = A\omega^2 = 0.04(158) = 6.32 m/s²

Solution

Worked Example: Wave Speed

Problem

A sound wave has f=440f = 440 Hz and λ=0.773\lambda = 0.773 m. Find the speed of sound.

v=fλ=440×0.773=340v = f\lambda = 440 \times 0.773 = 340 m/s

Solution

7. Practice Questions

    1. Define SHM and state the conditions required. (2 marks)
    1. A wave has speed 340 m/s and frequency 850 Hz. Calculate the wavelength. (1 mark)
    1. The amplitude of a wave doubles. By what factor does the intensity change? (1 mark)
    1. Distinguish between transverse and longitudinal waves with an example of each. (2 marks)

    Answers

    1. SHM: acceleration is proportional to displacement and directed towards a fixed equilibrium point. Conditions: restoring force proportional to displacement.

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Summary

  • SHM: a=ω2xa = -\omega^2 x; vmax=Aωv_{\max} = A\omega; amax=Aω2a_{\max} = A\omega^2
  • v=fλv = f\lambda; T=1/fT = 1/f; ω=2πf\omega = 2\pi f
  • Transverse (polarisable) vs longitudinal (compressions/rarefactions)
  • IA21/r2I \propto A^2 \propto 1/r^2 for point source

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