a submarine fitted with a SONAR system operates at a frequency 50kHz. An enwmy submarine is now moving towards it with a speed of 450km/hr. Find the frequency of the sound received back at the SONAR from the enemy submarine. Velocity of sound in water is 1450m/s.
the distance is closing, so the frequency is higher
F = 50kHz * [1450 / (1450 m/s - 450 km/hr)]
Thank you Sir. :)
To solve this problem, we can use the concept of Doppler effect. Doppler effect is the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave.
In this case, the submarine fitted with the SONAR system is the observer, and the enemy submarine is the source of the waves.
The formula for the Doppler effect with moving source is:
f' = (v + vr) / (v + vs) * f
Where:
f' = frequency observed by the observer
f = frequency emitted by the source
v = velocity of sound in the medium (given as 1450 m/s)
vr = velocity of the receiver (SONAR submarine)
vs = velocity of the source (enemy submarine)
We are given:
f = 50 kHz = 50,000 Hz
v = 1450 m/s
vs = 450 km/hr = 450,000 m/hr = 125 m/s (convert km/hr to m/s)
vr = 0 m/s (since the observer submarine is stationary)
Substituting these values into the formula:
f' = (v + vr) / (v + vs) * f
f' = (1450 + 0) / (1450 + 125) * 50,000
f' = 1450 / 1575 * 50,000
f' = 90.47% * 50,000
f' = 45235 Hz
Therefore, the frequency of the sound received back at the SONAR from the enemy submarine is 45,235 Hz.