A SONAR inside the sea works at 40 kHz. A submarine is approaching it with a velocity 360 Km $h^{- 1}$ . If the speed of sound in water is 1450 m $s^{- 1}$ then the apparent frequency of waves after reflection from submarine will be

11.5 kHz

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36.8 kHz

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42.9 kHz

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98.6 kHz

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Solution

The correct option is D 42.9 kHz
Velocity of submarine $V_{S} = 360 K m / h r = 360 * (1000 / 3600) m / s e c = 100 m s^{- 1}$

Velocity of Sound in Water $V = 1450 m s^{- 1}$

Frequency of SONAR $f = 40 k H z$

Applying Doppler's Effect in sound,

the apparant frequency $f^{'}$ by reciever will be,

$f^{'} = \frac{V}{V - V_{S}} f = \frac{1450}{1450 - 100} 40 k H z = \frac{1450 \times 40}{1350} k H z = 42.96 k H z$

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A SONAR inside the sea works at 40 kHz. A submarine is approaching it with a velocity 360 Kmh−1. If the speed of sound in water is 1450 ms−1 then the apparent frequency of waves after reflection from submarine will be

A SONAR inside the sea works at 40 kHz. A submarine is approaching it with a velocity 360 Km $h^{- 1}$ . If the speed of sound in water is 1450 m $s^{- 1}$ then the apparent frequency of waves after reflection from submarine will be