Q.
A source of sonic oscillations with frequency n = 1700 Hz and a receiver are located on the same normal to a wall. Both the source and receiver are stationary, and the wall recedes from the source with velocity u = 6.0 cm/s. Find the beat frequency (in Hz) registered by the receiver. The velocity of sound is equal to v = 340 m/s.
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
answer is 0.60.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
In this case the observer will notice two frequencies. One, direct through the source and second, through the reflection by moving wall.As both source and receiver are at rest the frequency of sound waves which directly reach to receiver, will be 1700 Hz.Hence ndirect =1700Hz. The frequency of sound which wall will receive as a moving observer isn1=n0v−uv−0=n0v−uv …(i)The wall now behaves as a moving source of frequency n1 which when received by receiver, the frequency observed is n2=n1v−0v−(−u)=n1vv+u=n0v−uv+u …(ii)=1700340−0.06340+0.06=1700×339.94340.06=1699.4HzThus, beat frequency received by detector is Δn=1700−1699.4=0.6Hz
Watch 3-min video & get full concept clarity