Question

A source of sound S emitting waves of frequency 100Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4m s⁻1 at an angle of 60^∘ with the source–observer line as shown in the figure. The observer is at rest. Find the apparent frequency observed by the observer. (Velocity of sound in air =330m s⁻1). 

• A. \(103\,\text{Hz}\)
• B. \(106\,\text{Hz}\)
• C. \(97\,\text{Hz}\)
• D. 100Hz

Hint:

In Doppler effect problems: vₛ = vcosθ Only the velocity component along the line of sight causes frequency change.

Answer 1

The Correct Option is A

Step 1: Only the component of source velocity along the line joining source and observer affects Doppler shift: vₛ = vcosθ = 19.4\cos60^∘ = 9.7m s⁻1
Step 2: Doppler formula for a moving source and stationary observer: f' = f((v)/(v - vₛ))
Step 3: Substituting values: f' = 100((330)/(330-9.7)) =100((330)/(320.3)) ≈ 103Hz