Doppler Effect MCQ - Practice Questions & Answers

JEE Main Syllabus 2025 PDF (Out) for Physics, Chemistry, Maths

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Doppler Effect is considered one the most difficult concept.
37 Questions around this concept.

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A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed $\nu$ ms^-1 The velocity of sound in air is 300 ms^-1. If the person can hear frequencies up to a maximum of 10000 Hz, the maximum value of $\nu$ upto which he can hear the whistle is :

$30ms^{-1}$

$15\sqrt{2}ms^{-1}$

$15/\sqrt{2}ms^{-1}$

$15 ms^{-1}$

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Doppler Effect

Doppler Effect -

Whenever there is a relative motion between a source of sound and the listener, the apparent frequency/wavelength of sound heard by the listener is different from the actual frequency/wavelength of sound emitted by the source.

When the distance between the source and listener is increasing the apparent frequency decreases. It means the apparent frequency is less than the actual frequency of sound. The reverse of this process is also true.

$\text { General expression for apparent frequency } f_{app}=\frac{\left[\left(v+v_{m}\right)-v_{o}\right] f}{\left[\left(v+v_{m}\right)-v_{s}\right]}$

Now, for different conditions the value of apparent frequency will change. Here f = Actual frequency; v₀= Velocity of observer; v_S = Velocity of source, v_M = Velocity of medium and v = Velocity of sound wave

There are some sign convention for the velocities - along the direction Source to Listener are taken as positive and all velocities along the direction Listener to Source are taken as negative.

If the velocity of the medium is zero then the formula become - $f_{app}=\left(\frac{v-v_{o}}{v-v_{s}}\right) f$

Now we will discuss some important cases and based on that the formulaes -

$\begin{array}{l}{\text { (1) Source is moving towards the listener and the listener at rest } f_{app}=\frac{v}{v-v_{s}}. f} \\ \\ {\text { (2) Source is moving away from the listener and the listener is at rest } f_{app}=\frac{v}{v+v_{s}}. f} \\ \\ {\text { (3) Source is at rest but listener is moving away from the source } f_{app}=\frac{v-v_{L}}{v} f}\end{array}$

$\begin{array}{l}{\text { (4) Source is at rest but the listener is moving towards the source } f_{app}=\frac{v+v_{L}}{v} f} \\ \\ {\text { (5) When the Source and listener are approaching each other } f_{app}=\left(\frac{v+v_{L}}{v-v_{S}}\right) f} \\ \\ {\text { (6) When the Source and listener moving away from each other } f_{app}=\left(\frac{v-v_{L}}{v+v_{s}}\right){f}}\end{array}$

Note - Source and listener moves perpendicular to the direction of wave propagation i.e., f_app = f. It means there is no change in frequency of sound heard for the small displacement of source and listener at right angle to the direction of wave propagation but this is not true for large displacement. For a large displacement the frequency decreases because the distance between source of sound and listener increases.