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Question
Two whistles A and B have frequencies 660 Hz and 590 Hz respectively. An observer is standing in the middle of the line joining the two sources. he middle of the line joining the two sources. Source B and observer are moving towards right with velocity 30 m/s and A is stationary at left side. The number of beats listened by the observer are ______.
(Velocity of sound in air = 300 m/s)
Options
2/s
4/s
6/s
8/s
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Solution
Two whistles A and B have frequencies 660 Hz and 590 Hz respectively. An observer is standing in the middle of the line joining the two sources. he middle of the line joining the two sources. Source B and observer are moving towards right with velocity 30 m/s and A is stationary at left side. The number of beats listened by the observer are 4/s.
Explanation:
The number of beats is the difference between the frequencies heard by the observer from the two source.
Frequency from source A:
Since, Observer is receding from the stationery source Apparent Frequency of source A:
\[\mathbf{f}^{\prime}=\mathbf{f}\left(\frac{\mathbf{V}-\mathbf{V}_0}{\mathbf{V}}\right)\]
\[=660\frac{ \begin{pmatrix} 300-30 \end{pmatrix}}{300}\]
= 594 Hz
Frequency of source B
Since, both observer and source B are moving with same velocity, Doppler effect won't happen
\[\begin{array} {cc}\mathrm{f_A^{^{\prime}}=f_B} & =590\mathrm{H_Z} \end{array}\]
\[\therefore\] No. of beats listened by the observer
\[=\mid\mathrm{f}_{\mathrm{A}}^{^{\prime}}-\mathrm{f}_{\mathrm{B}}^{^{\prime}}\mid\]
= |594 - 590|
= 4/s
