Doppler effect and beats

JEE Advanced 2019 Paper 1, Question 15

A train S1, moving with a uniform velocity of 108 km/h, approaches another train S2 standing on a platform. An observer O moves with a uniform velocity of 36 km/h towards S2, as shown in figure. Both the trains are blowing whistles of same frequency 120 Hz. When \mathrm{O} is 600 m away from S2 and distance between \mathrm{S} 1 and \mathrm{S} 2 is 800 m, what is the number of beats heard by O?

[Speed of the sound =330 m/s ]

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Recommended reading: How does Doppler effect work?


When there is relative motion between the source and observer, the observed frequency f_o is related to the source frequency f_s by

(1)   \begin{equation*}   f_O = \frac{v + v_O}{v-v_S} f_S ,  \end{equation*}

where v is the velocity of sound in the medium, v_o is the velocity with which the observer moves towards the source, and v_s is the velocity with which the source moves towards the observer. This is the equation for the Doppler Effect.

In the given problem both sources (the trains) emit sound with frequency 120 Hz. The observer O hears sounds of frequencies

(2)   \begin{align*}   f^{(1)}_O &= \frac{v+v_O \cos\theta}{v-v_{S1} \sin{\theta}} f_{S1} = \frac{336}{306} \times 120 \, {\rm Hz} , \\[1em]   f^{(2)}_O &= f_{S2} = \frac{v+v_O}{v} = \frac{340}{330} \times 120 \, {\rm Hz} \end{align*}

where the angle \theta is indicated in the figure below and we have plugged in v_O = 36 km/h, v_{S1} = 108 km/h, \cos \theta = \frac{3}{5}, and \sin \theta = \frac{4}{5}.

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The beat frequency is

(3)   \begin{equation*}   \boxed{f_{\rm beats} = \left| f^{(1)}_O-f^{(2)}_O \right| = 0.81 \ {\rm beats/second} .} \end{equation*}

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