MCQ
A body is moving forwards and backward. Change in frequency observed of source is 2%. What is velocity of the body? (Speed of sound is 300 m/s)
- A3 m/s
- B2.5 m/s
- C2 m/s
- D6 m/s
✓
Answer
$3 m / s$
Explanation: When the source is moving forward towards the observer, the apparent frequency is $f_1=\frac{v}{v-v_{ s }} \times f$
When source moves backwards $f_2=\frac{v}{v+v_s} \times f$
$
f_2-f_1=f v\left[\frac{1}{v+v_s}-\frac{1}{v-v_s}\right]=f v\left[\frac{-2 v_s}{v^2-v_s^2}\right]
$
As $v_{ s }<< v$, so
$
\begin{aligned}
& \frac{f_2-f_1}{f}=\left|\frac{2 v_s}{v}\right|=\frac{2}{100} \\
& v_s=\frac{v}{100}=\frac{300}{100}=3 m / s
\end{aligned}
$
Explanation: When the source is moving forward towards the observer, the apparent frequency is $f_1=\frac{v}{v-v_{ s }} \times f$
When source moves backwards $f_2=\frac{v}{v+v_s} \times f$
$
f_2-f_1=f v\left[\frac{1}{v+v_s}-\frac{1}{v-v_s}\right]=f v\left[\frac{-2 v_s}{v^2-v_s^2}\right]
$
As $v_{ s }<< v$, so
$
\begin{aligned}
& \frac{f_2-f_1}{f}=\left|\frac{2 v_s}{v}\right|=\frac{2}{100} \\
& v_s=\frac{v}{100}=\frac{300}{100}=3 m / s
\end{aligned}
$
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