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Wave Behaviour - The Doppler Effect

Grade 12IBPhysics

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

🔑Concepts

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The Doppler Effect is the change in the observed frequency f′f' of a wave when there is relative motion between the source and the observer.

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For sound waves, when a source moves towards a stationary observer, the wavefronts are compressed, leading to a shorter observed wavelength λ′\lambda' and a higher observed frequency f′f'.

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If the source moves away from a stationary observer, the wavefronts are stretched, leading to a longer observed wavelength λ′\lambda' and a lower observed frequency f′f'.

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For a stationary source and a moving observer, the frequency change is due to the observer intercepting wavefronts at a different rate (relative speed changes).

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In Electromagnetic (EM) waves, the Doppler Effect results in a 'redshift' (increase in wavelength) if the source moves away, and a 'blueshift' (decrease in wavelength) if it moves towards the observer.

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For EM waves where the relative speed vv is much less than the speed of light cc (v≪cv \ll c), the approximation Δff=Δλλ≈vc\frac{\Delta f}{f} = \frac{\Delta \lambda}{\lambda} \approx \frac{v}{c} is used.

📐Formulae

f′=f(vv∓vs) (Moving source, stationary observer)f' = f \left( \frac{v}{v \mp v_s} \right) \text{ (Moving source, stationary observer)}

f′=f(v±vobsv) (Moving observer, stationary source)f' = f \left( \frac{v \pm v_{obs}}{v} \right) \text{ (Moving observer, stationary source)}

Δf=vcf (Electromagnetic waves, v≪c)\Delta f = \frac{v}{c} f \text{ (Electromagnetic waves, } v \ll c \text{)}

z=Δλλ0≈vc (Redshift parameter)z = \frac{\Delta \lambda}{\lambda_0} \approx \frac{v}{c} \text{ (Redshift parameter)}

💡Examples

Problem 1:

An ambulance emits a siren at a frequency of f=400 Hzf = 400\text{ Hz} and travels at vs=30 m s−1v_s = 30\text{ m s}^{-1} towards a stationary observer. If the speed of sound is v=340 m s−1v = 340\text{ m s}^{-1}, calculate the frequency f′f' heard by the observer.

Solution:

Using the formula for an approaching source: f′=f(vv−vs)f' = f \left( \frac{v}{v - v_s} \right). Substituting the values: f′=400(340340−30)=400(340310)≈438.7 Hzf' = 400 \left( \frac{340}{340 - 30} \right) = 400 \left( \frac{340}{310} \right) \approx 438.7\text{ Hz}.

Explanation:

Since the source is approaching the observer, the observed frequency is higher than the emitted frequency because the denominator (v−vs)(v - v_s) is smaller than vv.

Problem 2:

A distant galaxy emits light with a spectral line of wavelength λ=656.3 nm\lambda = 656.3\text{ nm}. If the line is observed on Earth at a wavelength of λ′=660.0 nm\lambda' = 660.0\text{ nm}, determine the velocity vv at which the galaxy is moving relative to Earth.

Solution:

First, find Δλ=660.0−656.3=3.7 nm\Delta \lambda = 660.0 - 656.3 = 3.7\text{ nm}. Use the Doppler shift formula for light: Δλλ=vc\frac{\Delta \lambda}{\lambda} = \frac{v}{c}. Thus, v=c⋅Δλλ=(3×108 m s−1)⋅3.7656.3≈1.69×106 m s−1v = c \cdot \frac{\Delta \lambda}{\lambda} = (3 \times 10^8\text{ m s}^{-1}) \cdot \frac{3.7}{656.3} \approx 1.69 \times 10^6\text{ m s}^{-1}.

Explanation:

The observed wavelength is longer (redshifted), indicating that the galaxy is moving away from the observer. We use the non-relativistic approximation since vv is much smaller than cc.

The Doppler Effect - Revision Notes & Key Formulas | IB Grade 12 Physics