Question

The engine of a certain aircraft can produce a sound of 520 Hz. Suppose that one such aircraft is at rest on the runway while a second identical aircraft flies overhead at 80% the speed of sound. The pilot of each aircraft can listen to the sound produced by the engine of the other aircraft. Then the frequency heard by the pilot in the moving aircraft is (Assume the speed of sound in air as 320 m/s)
1. 544 Hz
2. 276 Hz
3. 104 Hz
4. 322 Hz

Answer 1

Correct Answer - Option 3 : 104 Hz

The correct answer is option 3) i.e. 104 Hz

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: Speed of sound (v) = 320 m/s and Frequency of source (f₀) = 520 Hz

Since the aircraft is at rest, Velocity of source, v_s = 0 m/s

• The velocity of moving aircraft,

⇒ v₀ = 80% × v = 0.8 × 320 = 256 m/s

• The frequency of the sound heard by the pilot in moving aircraft, 

\(⇒ f' = \frac{v - v_0}{v + v_s} f_0\)

\(⇒ f' = \frac{320 - 256}{320 + 0}\times 520 = 104 \: Hz\)