Question:
The velocity of sound in air is 330 $ms^{-1}$. To increase the apparent frequency of the sound by 50%, the source should move towards the stationary observer with a velocity equal to}
The velocity of sound in air is 330 $ms^{-1}$. To increase the apparent frequency of the sound by 50%, the source should move towards the stationary observer with a velocity equal to}
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If frequency becomes 1.5x, then $(v-v_{s})$ must be $v/1.5$.
Updated On: Jul 4, 2026
- 330 $ms^{-1}$
- 220 $ms^{-1}$
- 165 $ms^{-1}$
- 110 $ms^{-1}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The Doppler effect for a source moving towards a stationary observer is $f' = f \left( \frac{v}{v - v_{s}} \right)$.
Step 2: Meaning
Increasing frequency by 50% means $f' = 1.5f$ or $f' = \frac{3}{2}f$.
Step 3: Analysis
$\frac{3}{2}f = f \left( \frac{330}{330 - v_{s}} \right) \implies \frac{3}{2} = \frac{330}{330 - v_{s}} \implies 3(330 - v_{s}) = 660$.
Step 4: Conclusion
$990 - 3v_{s} = 660 \implies 3v_{s} = 330 \implies v_{s} = 110$ $ms^{-1}$. Final Answer: (D)
The Doppler effect for a source moving towards a stationary observer is $f' = f \left( \frac{v}{v - v_{s}} \right)$.
Step 2: Meaning
Increasing frequency by 50% means $f' = 1.5f$ or $f' = \frac{3}{2}f$.
Step 3: Analysis
$\frac{3}{2}f = f \left( \frac{330}{330 - v_{s}} \right) \implies \frac{3}{2} = \frac{330}{330 - v_{s}} \implies 3(330 - v_{s}) = 660$.
Step 4: Conclusion
$990 - 3v_{s} = 660 \implies 3v_{s} = 330 \implies v_{s} = 110$ $ms^{-1}$. Final Answer: (D)
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