Waves Ques 70
70. 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: 70.(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$
BETA
