Question

An observer is standing below a freely falling source of sound of frequency 900 Hz. The change in the frequency noticed by the observer after 3 seconds of free fall of the source is (speed of sound in air is 330 ms\(^{-1}\) and acceleration due to gravity = 10 ms\(^{-2}\))

• A. 990 Hz
• B. 818.18 Hz
• C. 81.82 Hz
• D. 90 Hz

Hint:

The time interval between maxima and minima in a beat pattern is half the beat period.

Answer 1

The Correct Option is C

When two progressive waves with frequencies \(128 \, \text{Hz}\) and \(124 \, \text{Hz}\) superpose, they produce a beat frequency of \(4 \, \text{Hz}\). The time between a maximum and adjacent minimum intensity, which is half the beat period, is calculated as follows: \[ T_{\text{beat}} = \frac{1}{f_{\text{beat}}} = \frac{1}{4} \, \text{seconds} \] \[ \Delta t = \frac{T_{\text{beat}}}{2} = \frac{1}{8} \, \text{seconds} \]