COMEDK2022PhysicsWaves and Sound
A bat emitting an ultrasonic wave of frequency 4.5 10^4 ~Hz at speed of 6 ~m / s between two parallel walls. The two frequencies heard by the bat will be
Options
- A4.67 10^4 ~Hz , 4.34 10^4 ~Hz
- B4.34 10^4 ~Hz , 4.67 10^4 ~Hz
- C4.5 10^4 ~Hz , 5.4 10^4 ~Hz
- D4.67 10^3 ~Hz , 4.34 10^4 ~Hz
Correct answer
A. 4.67 10^4 ~Hz , 4.34 10^4 ~Hz
Step-by-step solution
The frequency of the source is f₀ = 4.5 10^4 Hz . The speed of the bat is v_b = 6 m/s . The speed of sound in air is taken as v = 330 m/s . When the bat moves towards one wall, the wall acts as a stationary observer receiving the sound and then as a stationary source reflecting it back to the moving bat. The frequency f₁ heard by the bat from the wall it is moving towards is given by the Doppler effect formula for a moving observer and stationary source: f₁ = f₀ ( v + v_b v ) = 4.5 10^4 ( 330 + 6 330 ) = 4.5 10^4 (