Question

The figure shows the location of a source and detector at time $t=0$ . The source and detector are moving with velocities $v_S=5\hat{i}$ $m{s}^{-1}$ and $v_D=10\hat{j}$ $m{s}^{-1}$ respectively. The frequency of signals received by the detector at the moment when the source crosses the origin is (the frequency of the source is 100 Hz. velocity of sound $330m{s}^{-1}$ )

• A. $97Hz$
• B. $47Hz$
• C. $90Hz$
• D. $60Hz$

Answer 1

Answer: (A)

Time taken by 'S' to travel from S to 0 = $20s$
In this time detector goes from $D_1$ to $D_2$
The signal received by the detector at $t=20s$ is not that produced by the source at 'O' but produced from a position $S_1$ prior to 'O'
$\frac{r}{v}=\frac{d}{v_s}cos\theta =\frac{d}{r}=\frac{v_s}{v}=\frac{5}{330}=\frac{1}{66}$
$sin\theta =\sqrt{1-(\frac{1}{66}{)}^2}\approx 1$
$\Rightarrow f_D=f\left(\frac{v-v_{0}sin\theta }{v-v_{s}cos\theta }\right)$
$=100\left(\frac{330-10}{330-\frac{5}{66}}\right)=97Hz$