Question

A source of sound of frequency $660\text{ Hz}$ and an observer are moving towards each other with speeds of $31\text{ kmph}$ and $23\text{ kmph}$ respectively. If the wind blows with a speed of $5\text{ kmph}$ from observer towards the source, then the frequency of the sound heard by the observer is: (Speed of sound in air $=340\text{ ms}^{-1}$)

• A. $630\text{ Hz}$
• B. $660\text{ Hz}$
• C. $690\text{ Hz}$
• D. $720\text{ Hz}$

Hint:

Always identify the direction of sound propagation first. If wind blows opposite to the propagation direction, subtract wind speed from the speed of sound. If it blows along the propagation direction, add wind speed to the speed of sound.

Answer 1

The Correct Option is C

Step 1: Set up the Doppler picture with wind.
Source and observer approach each other, so the heard frequency rises. With wind present, sound travels in an effective medium, so we use \[ f' = f\left(\frac{v_{eff} + v_o}{v_{eff} - v_s}\right). \]
Step 2: Convert speeds to $ms^{-1}$.
Using $1\,kmph = \tfrac{5}{18}\,ms^{-1}$: source $v_s = 31\times\tfrac{5}{18} \approx 8.61\,ms^{-1}$, observer $v_o = 23\times\tfrac{5}{18} \approx 6.39\,ms^{-1}$, wind $v_w = 5\times\tfrac{5}{18} \approx 1.39\,ms^{-1}$.
Step 3: Fix the effective sound speed.
The wind blows from observer toward the source, opposite to the sound's travel from source to observer, so it reduces the effective speed: $v_{eff} = 340 - 1.39 \approx 338.6\,ms^{-1}$.
Step 4: Build the numerator and denominator.
Numerator $v_{eff} + v_o = 338.6 + 6.39 = 344.99$. Denominator $v_{eff} - v_s = 338.6 - 8.61 = 329.99$.
Step 5: Compute the apparent frequency.
$f' = 660\times \dfrac{345}{330} = 660\times 1.0454 \approx 690\,Hz$.
Step 6: Conclude.
Both motions raise the pitch and the wind makes only a small change, giving about $690\,Hz$, which is option (3).
\[ \boxed{f' \approx 690\,Hz} \]