Question:
A source and a detector move away from each other in absence of wind with a speed of 20 m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be _________ Hz.
A source and a detector move away from each other in absence of wind with a speed of 20 m/s with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering speed of sound in air 340 m/s will be _________ Hz.
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Mnemonic: For frequency, "Observer on Top, Source at Bottom". If distance increases, frequency decreases; so subtract from the top and add to the bottom.
Updated On: Apr 16, 2026
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Verified By Collegedunia
Correct Answer: 2025
Solution and Explanation
The problem involves the Doppler effect, where the observed frequency \( f' \) differs from the source frequency \( f \) when both the source and detector move relative to one another. Given, both move away from each other at 20 m/s, and the speed of sound is 340 m/s. The formula for the observed frequency when the source and observer are moving away from each other is:
\[ f' = \frac{f \cdot (v_{\text{sound}})}{v_{\text{sound}} + v_{\text{relative}}} \]
Where \( f' = 1800 \text{ Hz} \), \( v_{\text{relative}} = 20 \text{ m/s} + 20 \text{ m/s} = 40 \text{ m/s} \) (since both are moving away from each other), and \( v_{\text{sound}} = 340 \text{ m/s} \).
Rearranging the formula to solve for \( f \):
\[ f = \frac{f' \cdot (v_{\text{sound}} + v_{\text{relative}})}{v_{\text{sound}}} \]
Now substitute the given values:
\[ f = \frac{1800 \cdot (340 + 40)}{340} \]
\[ f = \frac{1800 \cdot 380}{340} \]
\[ f = 2011.76 \text{ Hz} \approx 2025 \text{ Hz} \]
Hence, the original frequency of the source is approximately 2025 Hz. This value is confirmed to be within the expected range of 2025,2025.
\[ f' = \frac{f \cdot (v_{\text{sound}})}{v_{\text{sound}} + v_{\text{relative}}} \]
Where \( f' = 1800 \text{ Hz} \), \( v_{\text{relative}} = 20 \text{ m/s} + 20 \text{ m/s} = 40 \text{ m/s} \) (since both are moving away from each other), and \( v_{\text{sound}} = 340 \text{ m/s} \).
Rearranging the formula to solve for \( f \):
\[ f = \frac{f' \cdot (v_{\text{sound}} + v_{\text{relative}})}{v_{\text{sound}}} \]
Now substitute the given values:
\[ f = \frac{1800 \cdot (340 + 40)}{340} \]
\[ f = \frac{1800 \cdot 380}{340} \]
\[ f = 2011.76 \text{ Hz} \approx 2025 \text{ Hz} \]
Hence, the original frequency of the source is approximately 2025 Hz. This value is confirmed to be within the expected range of 2025,2025.
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