Question

A train is approaching a stationary observer at a speed of 5 m/s with a whistle of frequency 100 Hz. Another train coming from the opposite direction approaches the stationary observer with a velocity 10 m/s and a whistle of frequency 120 Hz. If the velocity of sound in air is  320 m/s, then the beats heard by the observer will be

1. 22 Hz
2. 10 Hz
3. 54 Hz
4. 9 Hz

Answer 1

Correct Answer - Option 1 : 22 Hz

The correct answer is option 1) i.e. 22 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, 

​\(\Rightarrow f ' =\frac{v + v_0}{v + v_s} f_0\)

• 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, 

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

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.

• Beats: Beats are the periodic fluctuations heard in the intensity of a sound when two sound waves of nearly identical frequencies interfere with one another.
  • ​The number of beats is found from the difference in frequencies of two interfering waves. 

CALCULATION:

In the given question, both the trains are approaching the stationary observer.

• Since the observer is stationary, v₀ = 0 m/s
• Train 1 moving towards the observer: f₀ = 100 Hz, v_s = 5 m/s 

\(\Rightarrow f_1' =\frac{v + v_0}{v - v_s} f_0 = \frac{320 + 0}{320 - 5}\times 100 = 101.6 \approx 102\: Hz\)

• Train 1 moving towards the observer: f0 = 120 Hz, vs = 10 m/s

\(\Rightarrow f_2' =\frac{v + v_0}{v - v_s} f_0 = \frac{320 + 0}{320 - 10}\times 120 = 123.8 \approx 124\: Hz\)

• Beats = difference in frequencies = f₂' - f₁' = 124 - 102 = 22 Hz