Waves - Result Question 70
72. A source of sound S emitting waves of frequency $100 Hz$ and an observor $O$ are located at some distance from each other. The source is moving with a speed of $19.4 ms^{-1}$ at an angle of $60^{\circ}$ 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 $330 ms^{-1}$ )
[2015 RS]
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Answer:
Correct Answer: 72. (a)
Solution:
- (a) Here, original frequency of sound, $f_0=100$ $Hz$
Speed of source $V_s=19.4 \cos 60^{\circ}=9.7$
$19.4 \cos 60^{\circ}=9.7$
From Doppler’s formula
$f^{\prime}=f_0(\frac{V-V_0}{V-V_s})$
$f^{\prime}=100(\frac{V-0}{V-(+9.7)})$
$f^{\prime}=100 \frac{V}{V(1-\frac{9.7}{V})}=\frac{100}{(1-\frac{9.7}{330})}$
$=103 Hz$
Apparent frequency $ff=103 Hz$
