Question

A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to

• A. 6
• A. (d) 12
• B. 3
• C. 0

Answer 1

Answer: (A), (A)

6

Answer 2

App. Frequency due to train which is coming in $n_{1} = \frac{v}{v - v_{s}}.n$

App. Frequency due to train which is leaving $n_{2} = \frac{v}{v + v_{s}}.n$

So number of beats n₁ – n₂ = $\left( \frac{1}{316} - \frac{1}{324} \right)320 \times 240$

⇒ n₁ – n₂ = 6