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 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 $=340\text{ ms}^{-1}$)
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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.
Updated On: Jun 15, 2026
- $630\text{ Hz}$
- $660\text{ Hz}$
- $690\text{ Hz}$
- $720\text{ Hz}$
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The Correct Option is C
Solution and Explanation
Concept:
The apparent change in frequency of a sound wave due to the relative motion between the source and the observer is known as the Doppler Effect. When wind is present, the speed of sound relative to the ground changes, and its effect must also be incorporated.
For sound propagation in the presence of wind, the Doppler formula may be written as
\[
f' = f\left(\frac{v_{\text{eff}}+v_o}{v_{\text{eff}}-v_s}\right),
\]
where:
• $f$ = actual frequency emitted by the source
• $f'$ = apparent frequency heard by the observer
• $v_{\text{eff}}$ = effective speed of sound in the direction of propagation
• $v_o$ = speed of observer towards the source
• $v_s$ = speed of source towards the observer
Since frequency increases whenever source and observer move towards each other, both motions contribute to an increase in the observed frequency.
Step 1: Convert all given velocities from kmph to m/s. The standard conversion is \[ 1\text{ kmph}=\frac{5}{18}\text{ m/s}. \] For the source, \[ v_s=31\times\frac{5}{18} =\frac{155}{18}\text{ m/s}. \] For the observer, \[ v_o=23\times\frac{5}{18} =\frac{115}{18}\text{ m/s}. \] For the wind, \[ v_w=5\times\frac{5}{18} =\frac{25}{18}\text{ m/s}. \]
Step 2: Determine the effective velocity of sound. The wind is blowing from observer towards source. Therefore, the wind direction is opposite to the direction of sound propagation. Hence, \[ v_{\text{eff}}=v-v_w. \] Substituting values, \[ v_{\text{eff}} = 340-\frac{25}{18}. \]
Step 3: Apply the Doppler Effect formula. Since both source and observer move towards each other, \[ f' = 660 \left( \frac{(340-\frac{25}{18})+\frac{115}{18}} {(340-\frac{25}{18})-\frac{155}{18}} \right). \] Now simplify numerator: \[ 340+\frac{115-25}{18} = 340+\frac{90}{18} = 340+5 = 345. \] Similarly, \[ 340-\frac{25+155}{18} = 340-\frac{180}{18} = 340-10 = 330. \] Therefore, \[ f' = 660\left(\frac{345}{330}\right). \] \[ f' = 660\times\frac{23}{22}. \] \[ f' = 30\times23. \] \[ f' = 690\text{ Hz}. \]
Final Conclusion: The apparent frequency heard by the observer is \[ \boxed{690\text{ Hz}}. \] Hence, option (C) is correct.
• $f$ = actual frequency emitted by the source
• $f'$ = apparent frequency heard by the observer
• $v_{\text{eff}}$ = effective speed of sound in the direction of propagation
• $v_o$ = speed of observer towards the source
• $v_s$ = speed of source towards the observer
Since frequency increases whenever source and observer move towards each other, both motions contribute to an increase in the observed frequency.
Step 1: Convert all given velocities from kmph to m/s. The standard conversion is \[ 1\text{ kmph}=\frac{5}{18}\text{ m/s}. \] For the source, \[ v_s=31\times\frac{5}{18} =\frac{155}{18}\text{ m/s}. \] For the observer, \[ v_o=23\times\frac{5}{18} =\frac{115}{18}\text{ m/s}. \] For the wind, \[ v_w=5\times\frac{5}{18} =\frac{25}{18}\text{ m/s}. \]
Step 2: Determine the effective velocity of sound. The wind is blowing from observer towards source. Therefore, the wind direction is opposite to the direction of sound propagation. Hence, \[ v_{\text{eff}}=v-v_w. \] Substituting values, \[ v_{\text{eff}} = 340-\frac{25}{18}. \]
Step 3: Apply the Doppler Effect formula. Since both source and observer move towards each other, \[ f' = 660 \left( \frac{(340-\frac{25}{18})+\frac{115}{18}} {(340-\frac{25}{18})-\frac{155}{18}} \right). \] Now simplify numerator: \[ 340+\frac{115-25}{18} = 340+\frac{90}{18} = 340+5 = 345. \] Similarly, \[ 340-\frac{25+155}{18} = 340-\frac{180}{18} = 340-10 = 330. \] Therefore, \[ f' = 660\left(\frac{345}{330}\right). \] \[ f' = 660\times\frac{23}{22}. \] \[ f' = 30\times23. \] \[ f' = 690\text{ Hz}. \]
Final Conclusion: The apparent frequency heard by the observer is \[ \boxed{690\text{ Hz}}. \] Hence, option (C) is correct.
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