Wave Motion Ques 10
- A submarine $A$ travelling at $18 km / h$ is being chased along the line of its velocity by another submarine $B$ travelling at $27 km / h . B$ sends a sonar signal of $500 Hz$ to detect $A$ and receives a reflected sound of frequency $v$. The value of $v$ is close to (Speed of sound in water $=1500 ms^{-1}$ )
(2019 Main, 12 April I)
(a) $504 Hz$
(b) $507 Hz$
(c) $499 Hz$
(d) $502 Hz$
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Answer:
Correct Answer: 10.(d)
Solution:
- Given, velocity of submarine $(A)$,
$$ \begin{aligned} \qquad v _A & =18 km / h=\frac{18000}{3600} m / s \\ \text { or } \quad v _A & =5 m / s \end{aligned} $$
and velocity of submarine $(B)$,
$$ v _B=27 km / h=\frac{27000}{3600} m / s $$
or $\quad v _B=7.5 m / s$
Signal sent by submarine $(B)$ is detected by submarine $(A)$ can be shown as
Frequency of the signal, $f _o=500 Hz$
So, in this relative motion, frequency received by submarine $(A)$ is
$$ \begin{aligned} \quad f _1 & =\left(\frac{v _S-v _A}{v _S-v _B}\right) f _o=\left(\frac{1500-5}{1500-7.5}\right) 500 Hz \\ \Rightarrow \quad f _1 & =\frac{1495}{1492.5} \times 500 Hz \end{aligned} $$
The reflected frequency $f _1$ is now received back by submarine $(B)$.
So, frequency received at submarine $(B)$ is
$$ \begin{array}{rlrl} & f _2=\left(\frac{v _S+v _B}{v _S+v _A}\right) & f _1 & =\left(\frac{1500+7.5}{1500+5}\right)\left(\frac{1495}{1492.5}\right) 500 Hz \\ \Rightarrow & & f _2 & =\left(\frac{1507.5}{1505}\right)\left(\frac{1495}{1492.5}\right) 500 Hz \\ \Rightarrow & & f _2 & =1.00166 \times 1.00167 \times 500 \\ \Rightarrow & & f _2 & =501.67 Hz \\ & & \approx 502 Hz \end{array} $$
BETA
