Question

Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km h−1 and the other at 54 km h−1 relative to the water. The first submarine sends a sound signal (sound waves in water are also called sonar) at a frequency of 2000 Hz. (a) At what frequency is this signal received from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of of the sound wave in water to be 1500 m s−1.

Answer 1

Apparent frequency received by II submarine is 2034 Hz

Explanation:

• (a)The frequency of the sound, ν = 2000 Hz  

Speed of sound V = 1500 $ms^-^1$.    

The speed of the source $u_s$ = 36 $kmh^-^1$  

= $\frac {36000}{3600}$ $ms^-^1$

= 10 $ms^-^1$

The speed of the observer $u_0$ = 54 $kmh^-^1$  

= $\frac {54000}{3600}$ $ms^-^1$

= \frac {30}{2} $ms^-^1$ = 15 $ms^-^1$

• Hence the apparent frequency received by the second submarine  

• ν' = $\frac {(V + u_o)v}{(V+u_s)}$

 = $\frac {(1500+15)2000}{(1500-10)}$ 

= $1515 \times \frac {2000}{1490}$Hz  

= 2034 Hz  

• (b) This apparent frequency is now a source when the signal is reflected. The frequency of the sound received by the first submarine

ν" = $\frac {(V + u_o)v}{(V+u_s)}$

But now $u_0$ = 10 $ms^-^1$ and $u_s$ = 15 $ms^-^1$ and ν = ν' =2034 Hz  

Hence, ν"= $\frac {(1500+10)\times 2034}{(1500-15)}$

= $\frac {1510\times 2034}{1485}$

= 2068 Hz