Question:
A sound source of 1000 Hz frequency approaches the observer with speed \( 20~\text{ms}^{-1} \). The observed frequency of sound is nearly: (speed of sound in air \( = 340~\text{ms}^{-1} \))
A sound source of 1000 Hz frequency approaches the observer with speed \( 20~\text{ms}^{-1} \). The observed frequency of sound is nearly: (speed of sound in air \( = 340~\text{ms}^{-1} \))
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When a sound source moves towards you, the denominator must decrease (\( v - v_s \)), which naturally causes the fraction to be greater than 1, resulting in a higher apparent pitch. Keeping track of this physical behavior helps prevent mixing up sign conventions.
Updated On: Jun 8, 2026
- 1060 Hz
- 940 Hz
- 1020 Hz
- 1000 Hz
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The Correct Option is A
Solution and Explanation
Concept:
According to the Doppler Effect for sound waves, when a source of sound is moving directly towards a stationary observer, the apparent frequency \( f' \) heard by the observer increases and is given by the formula:
\[
f' = f \left( \frac{v}{v - v_s} \right)
\]
where \( f \) is the actual frequency, \( v \) is the speed of sound in air, and \( v_s \) is the speed of the moving source.
Step 1: Substituting the given values into the formula.
From the question, we have:
• Source frequency, \( f = 1000~\text{Hz} \)
• Speed of source, \( v_s = 20~\text{ms}^{-1} \)
• Speed of sound in air, \( v = 340~\text{ms}^{-1} \)
Plugging these parameters into our equation: \[ f' = 1000 \left( \frac{340}{340 - 20} \right) = 1000 \left( \frac{340}{320} \right) \]
Step 2: Calculating the apparent frequency value.
\[ f' = 1000 \times \frac{17}{16} = 1000 \times 1.0625 = 1062.5~\text{Hz} \approx 1060~\text{Hz} \] Thus, the observed frequency is approximately \( 1060~\text{Hz} \).
Step 1: Substituting the given values into the formula.
From the question, we have:
• Source frequency, \( f = 1000~\text{Hz} \)
• Speed of source, \( v_s = 20~\text{ms}^{-1} \)
• Speed of sound in air, \( v = 340~\text{ms}^{-1} \)
Plugging these parameters into our equation: \[ f' = 1000 \left( \frac{340}{340 - 20} \right) = 1000 \left( \frac{340}{320} \right) \]
Step 2: Calculating the apparent frequency value.
\[ f' = 1000 \times \frac{17}{16} = 1000 \times 1.0625 = 1062.5~\text{Hz} \approx 1060~\text{Hz} \] Thus, the observed frequency is approximately \( 1060~\text{Hz} \).
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