Question

While driving at a speed of 40 km/h, you notice your friend standing in a bus stop ahead of you. You press the horn before passing the bus stop to let him notice you. If the horn has a frequency of 150 Hz, the sound heard by your friend will have a frequency of (Assume the speed of sound in air as 320 m/s)
1. 305 Hz
2. 155 Hz
3. 230 Hz
4. 80 Hz

Answer 1

Correct Answer - Option 2 : 155 Hz

The correct answer is option 2) i.e. 155 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:

Frequency of horn, f0 = 150 Hz

• The velocity of sound,

⇒ vs = 40 km/h = 40 × (5/18) = 11.12 m/s

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

• The horn is pressed before reaching the bus stop and therefore, the source of sound is moving towards the observer at that instant.
• The frequency of the horn heard by the friend, 

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

\(⇒ f' = \frac{v + v_0}{v - v_s} f_0 = \frac{320 + 0}{320 - 11.12} \times 150 = 155.3 \approx 155 \:Hz\)