1.

Two submarines are approaching each other in a calm sea. The first submarine travels at a speed of 36 km/h and the other at 54 km/h 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 the signal received by the second submarine? (b) The signal is reflected from the second submarine. At what frequency is this signal received by the first submarine. Take the speed of the sound wave in water to be 1500 m/s.

Answer»

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

Speed of sound V = 1500 m/s.  

The speed of the source uₛ = 36 km/h 

=36000/3600 m/s =10 m/s 

The speed of the observer uₒ = 54 km/h 

=54000/3600 m/s 

=30/2 m/s =15 m/s 

Hence the apparent frequency received by the second submarine

 ν' = (V+uₒ)ν/(V-uₛ) 

=(1500+15)2000/(1500-10) 

=1515*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" =(V+uₒ)ν/(V-uₛ) 

But now uₒ = 10 m/s and uₛ = 15 m/s and ν = ν' =2034 Hz 

Hence, ν"=(1500+10)*2034/(1500-15) 

=1510*2034/1485 

=2068 Hz



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