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Question

Physics Question on doppler effect

A source of sound SS emitting waves of frequency 100Hz100\, Hz and an observor OO are located at some distance from each other. The source is moving with a speed of 19.4ms119.4\, m s^{-1} at an angle of 6060^{\circ} with the source observer line as shown in the figure. The observor is at rest. The apparent frequency observed by the observer is (velocity of sound in air 330ms1330\, ms ^{-1} )

A

103 Hz

B

106 Hz

C

97 Hz

D

100 Hz

Answer

103 Hz

Explanation

Solution

Here, original frequency of sound, f0=100Hzf_{0}=100 \,Hz Speed of source Vs19.4cos60=9.7V_{s} 19.4 \cos 60^{\circ}=9.7 From Doppler's formula f1=f0(VV0VVs)f^{1}=f_{0}\left(\frac{V-V_0}{V-V_{s}}\right) f1=100(V0V(+9.7))f^{1}=100\left(\frac{V-0}{V-(+9.7)}\right) f1100VV(19.7V)f^{1} 100 \frac{V}{V\left(1-\frac{9.7}{V}\right)} f1=100(1+9.7330)=103Hzf^{1}=100\left(1+\frac{9.7}{330}\right)=103 \,Hz Apparent frequency f1=103Hzf^{1}=103 \,Hz