Question

When a sound source of frequency n is approaching a stationary observer with velocity u, then the apparent change in frequency is Δn1 and when the same source is receding with velocity u from the stationary observer then the apparent change in frequency is
 Δn2 Then: 
1.  Δn1 =  Δn₂ = 0
2.  Δn1 =  Δn2 ≠  0
3.  Δn1 >  Δn2 
4.  Δn1 <  Δn2 

Answer 1

Correct Answer - Option 3 :  Δn1 >  Δn2 

The correct answer is option 3) i.e.  Δn1 >  Δn2 

CONCEPT:

• Doppler Effect: The phenomenon of change in frequency observed when a source of the sound and a listener are moving relative to each other is called the Doppler effect.
  • When the listener and the source are moving away from one another, the sound heard by the listener will have a frequency lower than the frequency of the sound from the source. 

The frequency of the sound heard by the listener, f' = \(\frac{v - v_0}{v + v_s} f_0\)

• When the listener and source are moving towards one another, the sound heard by the listener will have a frequency higher than the frequency of sound from the source.

The frequency of the sound heard by the listener, f' = \(\frac{v + v_0}{v - v_s} f_0\)

Where v is the velocity of sound in air, v0 is the velocity of the listener, vs is the velocity of the source of the sound, and f0 is the frequency of sound emitted by the source.

CALCULATION:

Given that:

Frequency of sound, f₀ = n

The velocity of sound, v_s = u

Since the observer is stationary, the velocity of the listener, v0 = 0 m/s

Case 1 - When the sound is approaching the listener

The change in frequency, f' = Δn1 = \(\frac{v + v_0}{v - v_s} f_0 = \frac{320 + 0}{320 - u} n = \frac{320n}{320 - u} \)      ----(1)

Case 2 - When the sound s moving away from the listener

The change in frequency, f' = Δn2 = \(\frac{v + v_0}{v + v_s} f_0 = \frac{320 + 0}{320 + u} n = \frac{320 n}{320 + u}\)       ----(2)

From (1) and (2)

Δn1 > Δn2