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Wave Behaviour - The Doppler Effect

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Doppler effect is defined as the change in the observed frequency (f′f') of a wave when there is relative motion between the source of the wave and the observer.

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When the source and observer move towards each other, the observed frequency (f′f') is higher than the emitted frequency (ff). This is due to the 'compression' of wavefronts in front of the moving source.

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When the source and observer move away from each other, the observed frequency (f′f') is lower than the emitted frequency (ff). This is due to the 'stretching' of wavefronts behind the moving source.

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For sound waves, the shift depends on whether the source or the observer is moving relative to the medium. For light (electromagnetic waves), the shift depends only on the relative velocity vv because the speed of light cc is constant in all inertial frames.

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In astronomy, a shift towards longer wavelengths (lower frequency) is called 'redshift', indicating the object is moving away. A shift towards shorter wavelengths (higher frequency) is called 'blueshift', indicating the object is moving closer.

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The approximation for electromagnetic waves Δff=Δλλ≈vc\frac{\Delta f}{f} = \frac{\Delta \lambda}{\lambda} \approx \frac{v}{c} is only valid when the relative velocity vv is much smaller than the speed of light (v≪cv \ll c).

📐Formulae

f′=f(vv∓vs)f' = f \left( \frac{v}{v \mp v_s} \right) (Moving Source: −- for approaching, ++ for receding)

f′=f(v±vov)f' = f \left( \frac{v \pm v_o}{v} \right) (Moving Observer: ++ for approaching, −- for receding)

Δf=vcf\Delta f = \frac{v}{c}f (Doppler shift for light)

z=Δλλ0≈vcz = \frac{\Delta \lambda}{\lambda_0} \approx \frac{v}{c} (Redshift ratio for v≪cv \ll c)

💡Examples

Problem 1:

An ambulance emitting a siren tone of f=1200 Hzf = 1200\text{ Hz} is traveling at vs=40 m s−1v_s = 40\text{ m s}^{-1} away from a stationary observer. If the speed of sound in air is v=340 m s−1v = 340\text{ m s}^{-1}, calculate the frequency f′f' heard by the observer.

Solution:

f′=f(vv+vs)=1200(340340+40)=1200(340380)≈1073.68 Hzf' = f \left( \frac{v}{v + v_s} \right) = 1200 \left( \frac{340}{340 + 40} \right) = 1200 \left( \frac{340}{380} \right) \approx 1073.68\text{ Hz}

Explanation:

Since the source is moving away from a stationary observer, we use the formula for a moving source with the plus sign in the denominator to account for the increase in wavelength and subsequent decrease in observed frequency.

Problem 2:

A distant galaxy has a known hydrogen emission line at λ0=656.3 nm\lambda_0 = 656.3\text{ nm}. However, the observed wavelength on Earth is λ=675.4 nm\lambda = 675.4\text{ nm}. Calculate the recessional velocity vv of the galaxy.

Solution:

Δλ=675.4−656.3=19.1 nm\Delta \lambda = 675.4 - 656.3 = 19.1\text{ nm}. Using v≈cΔλλ0v \approx c \frac{\Delta \lambda}{\lambda_0}, we get v=(3.0×108)(19.1656.3)≈8.73×106 m s−1v = (3.0 \times 10^8) \left( \frac{19.1}{656.3} \right) \approx 8.73 \times 10^6\text{ m s}^{-1}

Explanation:

The change in wavelength Δλ\Delta \lambda is calculated first. Since the observed wavelength is longer (redshifted), the galaxy is moving away. We use the electromagnetic Doppler approximation to find the velocity relative to the speed of light cc.

The Doppler Effect - Revision Notes & Key Formulas | IB Grade 11 Physics