### 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]

## Show Answer

#### 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 $ff=103 Hz$