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Oscillations and Waves - Doppler Effect

Grade 11ICSEPhysics

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

🔑Concepts

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The Doppler Effect is the apparent change in the frequency of a wave (sound or light) caused by the relative motion between the source of the wave and the observer.

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When the source and the observer move toward each other, the observed frequency f′f' is higher than the actual frequency ff.

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When the source and the observer move away from each other, the observed frequency f′f' is lower than the actual frequency ff.

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The phenomenon is observed in sound waves, light waves (Red shift and Blue shift), and electromagnetic waves like Radar.

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If only the observer moves, the wavelength of the sound in the medium remains unchanged, but the observer intercepts more or fewer wave fronts per second.

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If the source moves, the actual wavelength of the sound in the medium changes (λ′<λ\lambda' < \lambda when approaching, λ′>λ\lambda' > \lambda when receding).

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The Doppler Effect is not observed if the source and observer move in the same direction with the same velocity (relative velocity is zero), or if the motion is perpendicular to the line joining them.

📐Formulae

f′=f(v±vov∓vs)f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) where vv is velocity of sound, vov_o is velocity of observer, and vsv_s is velocity of source.

f′=f(v+vov)f' = f \left( \frac{v + v_o}{v} \right) (Observer moving towards stationary source)

f′=f(v−vov)f' = f \left( \frac{v - v_o}{v} \right) (Observer moving away from stationary source)

f′=f(vv−vs)f' = f \left( \frac{v}{v - v_s} \right) (Source moving towards stationary observer)

f′=f(vv+vs)f' = f \left( \frac{v}{v + v_s} \right) (Source moving away from stationary observer)

λ′=v−vsf\lambda' = \frac{v - v_s}{f} (Apparent wavelength when source moves towards observer)

💡Examples

Problem 1:

A train whistle emits a sound of frequency 400 Hz400\text{ Hz}. If the train is approaching a stationary observer at a speed of 20 m/s20\text{ m/s} and the speed of sound in air is 340 m/s340\text{ m/s}, calculate the apparent frequency heard by the observer.

Solution:

f′=f(vv−vs)f' = f \left( \frac{v}{v - v_s} \right) Given f=400 Hzf = 400\text{ Hz}, v=340 m/sv = 340\text{ m/s}, vs=20 m/sv_s = 20\text{ m/s}, vo=0v_o = 0. f′=400(340340−20)=400(340320)=425 Hzf' = 400 \left( \frac{340}{340 - 20} \right) = 400 \left( \frac{340}{320} \right) = 425\text{ Hz}

Explanation:

Since the source is moving towards the observer, the waves are 'compressed', leading to a decrease in wavelength and an increase in the observed frequency.

Problem 2:

An observer is moving away from a stationary siren at a speed of 10 m/s10\text{ m/s}. If the siren emits a frequency of 1000 Hz1000\text{ Hz} and the speed of sound is 330 m/s330\text{ m/s}, find the frequency heard.

Solution:

f′=f(v−vov)f' = f \left( \frac{v - v_o}{v} \right) Given f=1000 Hzf = 1000\text{ Hz}, v=330 m/sv = 330\text{ m/s}, vo=10 m/sv_o = 10\text{ m/s}, vs=0v_s = 0. f′=1000(330−10330)=1000(320330)≈969.7 Hzf' = 1000 \left( \frac{330 - 10}{330} \right) = 1000 \left( \frac{320}{330} \right) \approx 969.7\text{ Hz}

Explanation:

As the observer moves away, they encounter fewer wave crests per unit time, resulting in a lower perceived frequency.

Doppler Effect - Revision Notes & Key Formulas | ICSE Class 11 Physics