IB — physicsHard3 min read

Doppler Effect

Frequency shift for moving sources and observers; applications to sound and light

Tutor AI Team
Tutor AI Team

# Doppler Effect — IB Physics

The Doppler effect is the change in observed frequency (or wavelength) of a wave when the source and/or observer are moving relative to each other. It applies to sound, light, and all waves.

1. The Doppler Effect for Sound

Moving Source

f′=fvv±vsf' = f \frac{v}{v \pm v_s}

  • Source approaching: use v−vsv - v_s (denominator smaller → f′f' higher)
  • Source receding: use v+vsv + v_s (denominator larger → f′f' lower)

Moving Observer

f′=fv±vovf' = f \frac{v \pm v_o}{v}

  • Observer approaching: use v+vov + v_o → higher f′f'
  • Observer receding: use v−vov - v_o → lower f′f'

Where vv = speed of wave in medium, vsv_s = source speed, vov_o = observer speed.

2. The Doppler Effect for Light

Δff=Δλλ≈vc\frac{\Delta f}{f} = \frac{\Delta\lambda}{\lambda} \approx \frac{v}{c}

(Valid for v≪cv \ll c)

Red shift: Source moving away → wavelength increases (shifted towards red). z=Δλ/λz = \Delta\lambda/\lambda.

Blue shift: Source approaching → wavelength decreases (shifted towards blue).

3. Applications

  • Police radar guns: Detect speed via Doppler shift of reflected microwaves
  • Medical ultrasound: Measure blood flow speed
  • Astronomy: Red shift of galaxies → evidence for expanding universe
  • Weather radar: Detect wind speeds

Worked Example: Moving Source

Problem

An ambulance siren emits at 800 Hz. Speed of sound = 340 m/s. Ambulance at 25 m/s.

Approaching: f′=800×340/(340−25)=800×340/315=863f' = 800 \times 340/(340-25) = 800 \times 340/315 = 863 Hz Receding: f′=800×340/(340+25)=800×340/365=745f' = 800 \times 340/(340+25) = 800 \times 340/365 = 745 Hz

Solution

Worked Example: Red Shift

Problem

A galaxy has a hydrogen line at 660 nm instead of the laboratory value of 656 nm.

Δλ=660−656=4\Delta\lambda = 660 - 656 = 4 nm z=Δλ/λ=4/656=0.0061z = \Delta\lambda/\lambda = 4/656 = 0.0061 v=zc=0.0061×3×108=1.83×106v = zc = 0.0061 \times 3 \times 10^8 = 1.83 \times 10^6 m/s (away from Earth)

Solution

5. Practice Questions

    1. A car horn emits at 400 Hz. The car approaches at 30 m/s. Find the observed frequency (vsound=340v_{\text{sound}} = 340 m/s). (2 marks)
    1. A star shows a spectral line at 502 nm instead of 500 nm. Calculate the recession speed. (3 marks)
    1. Explain the Doppler effect for sound when a source moves towards an observer. (3 marks)

    Answers

    1. f′=400×340/(340−30)=400×340/310=439f' = 400 \times 340/(340-30) = 400 \times 340/310 = 439 Hz.
    1. z=Δλ/λ=2/500=0.004z = \Delta\lambda/\lambda = 2/500 = 0.004. v=0.004×3×108=1.2×106v = 0.004 \times 3 \times 10^8 = 1.2 \times 10^6 m/s.

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Summary

  • Source approaching: higher frequency (shorter wavelength)
  • Source receding: lower frequency (longer wavelength)
  • Sound: f′=fv/(v±vs)f' = fv/(v \pm v_s) or f′=f(v±vo)/vf' = f(v \pm v_o)/v
  • Light: v/c≈Δλ/λv/c \approx \Delta\lambda/\lambda (red shift = moving away)
Tags:
#ib#physics#waves#doppler-effect#frequency-shift#red-shift

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