A SONAR system fixed in a submarine operates at a frequency 40 kHz. An enemy submarine moves towards the SONAR with a speed of 360 km h-1. What is the frequency of sound reflected by the submarine? Take the speed of sound in water to be 1450 m s-1.


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Frequency of SONAR (source)

n = 40 kHz = 40 × 103 Hz

Speed of sound waves, v = 1450 m s-1

Speed of the listener, vL = 360 km h-1

= 360 × \(\frac{5}{18}\) ms-1

= 100 m s-1

Since, the source is at rest and the observer moves towards the source (SONAR).

We have,

n = n0 \((\frac{v+v_L}{v})\) = 40 × 103 × \((\frac{1450+100}{14540})\)

∴ n = 4.276 × 104 Hz

This frequency n’ is reflected by the enemy ship and is observed by the SONAR (which now acts as observer).

Therefore, in this case

vs = 100 m s-1

Apparent frequency,

n = n0 \((\frac{v}{v-v_s})\)

= 4.276 × 104 × \((\frac{1450}{1450-100})\) = 4.59 × 104 Hz

∴ n = 45.9 kHz

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