Question

A submarine (A) travelling at 18 \[km/hr\] is being chased along the line of its velocity by another submarine(B) travelling at 27 \[km/hr\]. B sends a sonar signal of 500 $Hz$ to detect A and receives a reflected sound of frequency $\upsilon $. The value of $\upsilon $ is close to: (Speed of sound in water is $1500m{s^{ - 1}}$
A) $499Hz$
B) $502Hz$
C) $507Hz$
D) $504Hz$

Answer 1

Hint: This question is based on the concept of Doppler’s effect.
The Doppler effect is the shift in frequency of a wave in relation to an observer who is moving relative to the wave source.

Complete step by step solution:
The Doppler’s effect formula for the shift in frequency is given by the formula:
$\upsilon = \left\{ {\dfrac{{c \pm {v_r}}}{{c \pm {v_s}}}} \right\}{\upsilon _0}$
where
$\upsilon $= observed frequency by the receiver
$c$= speed of sound in the medium
${v_r}$= velocity of the receiver
${v_s}$= velocity of the source
${\upsilon _0}$= emitted frequency
In this problem, we have two submarines moving in the same line with different speeds.

Here,
Receiver velocity (A), ${v_r} = 18km/hr = \dfrac{5}{{18}} \times 18 = 5m{s^{ - 1}}$
Sender velocity (B), ${v_s} = 27km/hr = \dfrac{5}{{18}} \times 27 = 7.5m{s^{ - 1}}$
Speed of sound in water, $c = 1500m{s^{ - 1}}$
Emitted frequency, ${\upsilon _0} = 500Hz$
Substituting the values in the Doppler effect equation, we get:
$
  \upsilon = \left\{ {\dfrac{{c \pm {v_r}}}{{c \pm {v_s}}}} \right\}{\upsilon _0} \\
  \upsilon = \left\{ {\dfrac{{1500 + 5}}{{1500 + 7.5}}} \right\}500 \\
  Solving, \\
  \upsilon = \left\{ {\dfrac{{1505}}{{1507.5}}} \right\}500 \\
  \upsilon = 0.9983 \times 500 \\
  \upsilon = 499.17Hz \approx 499Hz \\
 $
$\therefore$ The value of $\upsilon $ is close to $499Hz$. Hence, the correct option is Option A.

Note:
The signs of ${v_r}$ and ${v_s}$ depend on the relative movement between the source and the receiver.
If the receiver is moving towards the source, it is positive and if it is moving away, it is taken as negative.
Similarly,
If the source is moving towards the receiver, it is positive and if it is moving away, it is taken as negative.
Also, the frequency will always be lower than the emitted frequency if the receiver is moving away from the source. Hence, you can check the validity of your answer wherein if you get a higher frequency when you are moving away from the source, it is evident that you have gone wrong in calculations.