Question:
A source of sound moves towards a stationary observer with speed \( 10 \text{ m/s} \). Speed of sound = \( 340 \text{ m/s} \). If original frequency is \( 340 \text{ Hz} \), apparent frequency is:
A source of sound moves towards a stationary observer with speed \( 10 \text{ m/s} \). Speed of sound = \( 340 \text{ m/s} \). If original frequency is \( 340 \text{ Hz} \), apparent frequency is:
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To remember the sign convention for the Doppler Effect easily, think of distance: if the relative motion reduces the distance between the source and observer (approaching), the frequency must increase—meaning you subtract in the denominator or add in the numerator!
Updated On: May 20, 2026
- \( 350 \text{ Hz} \)
- \( 360 \text{ Hz} \)
- \( 370 \text{ Hz} \)
- \( 380 \text{ Hz} \)
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The Correct Option is A
Solution and Explanation
Concept:
The apparent change in the frequency of a wave caused by the relative motion between the source of the wave and the observer is known as the Doppler Effect.
The generalized master formula for the apparent frequency \( f' \) heard by an observer is given by:
\[
f' = f \cdot \left( \frac{v \pm v_o}{v \mp v_s} \right)
\]
where:
• \( f \) is the original (emitted) frequency of the sound source.
• \( v \) is the speed of sound waves in the medium.
• \( v_o \) is the velocity of the observer relative to the medium.
• \( v_s \) is the velocity of the source relative to the medium. When a source moves towards a stationary observer, the sound waves are compressed, causing the apparent frequency to be higher than the original frequency. The formula simplifies to: \[ f' = f \cdot \left( \frac{v}{v - v_s} \right) \]
Step 1: Extracting the given parameters from the problem text.
From the question statement, we have the following parameters:
• Original frequency of the source, \( f = 340 \text{ Hz} \)
• Speed of sound in air, \( v = 340 \text{ m/s} \)
• Speed of the moving source, \( v_s = 10 \text{ m/s} \)
• The observer is stationary, so \( v_o = 0 \text{ m/s} \)
Step 2: Substituting the values into the specialized Doppler effect equation.
Apply the values directly into the simplified formula: \[ f' = 340 \cdot \left( \frac{340}{340 - 10} \right) \] Evaluating the subtraction in the denominator: \[ 340 - 10 = 330 \text{ m/s} \] Now, substituting this back: \[ f' = 340 \cdot \left( \frac{340}{330} \right) = 340 \cdot \frac{34}{33} \]
Step 3: Calculating the final numerical value.
Let us carry out the arithmetic division and multiplication: \[ f' = \frac{11560}{33} \approx 350.3 \text{ Hz} \] Rounding off to the nearest whole number matching the multiple-choice options gives: \[ f' \approx 350 \text{ Hz} \] This precisely matches option (A).
• \( f \) is the original (emitted) frequency of the sound source.
• \( v \) is the speed of sound waves in the medium.
• \( v_o \) is the velocity of the observer relative to the medium.
• \( v_s \) is the velocity of the source relative to the medium. When a source moves towards a stationary observer, the sound waves are compressed, causing the apparent frequency to be higher than the original frequency. The formula simplifies to: \[ f' = f \cdot \left( \frac{v}{v - v_s} \right) \]
Step 1: Extracting the given parameters from the problem text.
From the question statement, we have the following parameters:
• Original frequency of the source, \( f = 340 \text{ Hz} \)
• Speed of sound in air, \( v = 340 \text{ m/s} \)
• Speed of the moving source, \( v_s = 10 \text{ m/s} \)
• The observer is stationary, so \( v_o = 0 \text{ m/s} \)
Step 2: Substituting the values into the specialized Doppler effect equation.
Apply the values directly into the simplified formula: \[ f' = 340 \cdot \left( \frac{340}{340 - 10} \right) \] Evaluating the subtraction in the denominator: \[ 340 - 10 = 330 \text{ m/s} \] Now, substituting this back: \[ f' = 340 \cdot \left( \frac{340}{330} \right) = 340 \cdot \frac{34}{33} \]
Step 3: Calculating the final numerical value.
Let us carry out the arithmetic division and multiplication: \[ f' = \frac{11560}{33} \approx 350.3 \text{ Hz} \] Rounding off to the nearest whole number matching the multiple-choice options gives: \[ f' \approx 350 \text{ Hz} \] This precisely matches option (A).
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