Question:
A source of sound and an observer are moving each with a speed of $10\%$ of the speed of sound in air. If the frequency of sound heard by the observer when they approach each other is $400\text{ Hz}$ more than the frequency of sound heard by the observer when they move away from each other, then the frequency of the source of sound is:
A source of sound and an observer are moving each with a speed of $10\%$ of the speed of sound in air. If the frequency of sound heard by the observer when they approach each other is $400\text{ Hz}$ more than the frequency of sound heard by the observer when they move away from each other, then the frequency of the source of sound is:
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When $v_o = v_s = u$, the ratio of approaching to receding frequency is:
$\frac{f_{\text{app}}}{f_{\text{rec}}} = \left(\frac{v+u}{v-u}\right)^2$.
Here, with $u=0.1v$, the ratio is $\left(\frac{1.1}{0.9}\right)^2 = \frac{121}{81}$.
This structural identity helps to quickly verify the common denominator $99$ in calculations.
$\frac{f_{\text{app}}}{f_{\text{rec}}} = \left(\frac{v+u}{v-u}\right)^2$.
Here, with $u=0.1v$, the ratio is $\left(\frac{1.1}{0.9}\right)^2 = \frac{121}{81}$.
This structural identity helps to quickly verify the common denominator $99$ in calculations.
Updated On: Jul 22, 2026
- $440\text{ Hz}$
- $770\text{ Hz}$
- $550\text{ Hz}$
- $990\text{ Hz}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
This question involves the Doppler effect in sound.
A source and an observer are moving with known relative speeds. We are given the difference in apparent frequencies between their approaching and receding states and must find the true frequency of the source.
Step 2: Key Formula and Approach:
The general Doppler formula for apparent frequency $f'$ is:
\[ f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \] where:
- $v$ is the speed of sound in air.
- $v_o$ is the speed of the observer.
- $v_s$ is the speed of the source.
- $f$ is the original source frequency.
Step 3: Detailed Explanation:
• Identify given speeds:
Let $v$ be the speed of sound in air.
Speed of source: $v_s = 0.1 v$
Speed of observer: $v_o = 0.1 v$
• Case 1: Approaching each other:
Both movements increase the frequency, so the observer moves towards the source (+ numerator) and the source moves towards the observer (- denominator):
\[ f_{\text{app}} = f \left( \frac{v + v_o}{v - v_s} \right) = f \left( \frac{v + 0.1v}{v - 0.1v} \right) = f \left( \frac{1.1v}{0.9v} \right) = \frac{11}{9}f \]
• Case 2: Moving away from each other:
Both movements decrease the frequency:
\[ f_{\text{rec}} = f \left( \frac{v - v_o}{v + v_s} \right) = f \left( \frac{v - 0.1v}{v + 0.1v} \right) = f \left( \frac{0.9v}{1.1v} \right) = \frac{9}{11}f \]
• Set up the frequency difference equation:
We are given $f_{\text{app}} - f_{\text{rec}} = 400\text{ Hz}$:
\[ \left( \frac{11}{9} - \frac{9}{11} \right) f = 400 \] \[ \left( \frac{121 - 81}{99} \right) f = 400 \] \[ \frac{40}{99} f = 400 \] \[ f = \frac{400 \times 99}{40} = 10 \times 99 = 990\text{ Hz} \]
Step 4: Final Answer:
The frequency of the source of sound is $990\text{ Hz}$, which corresponds to Option (D).
This question involves the Doppler effect in sound.
A source and an observer are moving with known relative speeds. We are given the difference in apparent frequencies between their approaching and receding states and must find the true frequency of the source.
Step 2: Key Formula and Approach:
The general Doppler formula for apparent frequency $f'$ is:
\[ f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) \] where:
- $v$ is the speed of sound in air.
- $v_o$ is the speed of the observer.
- $v_s$ is the speed of the source.
- $f$ is the original source frequency.
Step 3: Detailed Explanation:
• Identify given speeds:
Let $v$ be the speed of sound in air.
Speed of source: $v_s = 0.1 v$
Speed of observer: $v_o = 0.1 v$
• Case 1: Approaching each other:
Both movements increase the frequency, so the observer moves towards the source (+ numerator) and the source moves towards the observer (- denominator):
\[ f_{\text{app}} = f \left( \frac{v + v_o}{v - v_s} \right) = f \left( \frac{v + 0.1v}{v - 0.1v} \right) = f \left( \frac{1.1v}{0.9v} \right) = \frac{11}{9}f \]
• Case 2: Moving away from each other:
Both movements decrease the frequency:
\[ f_{\text{rec}} = f \left( \frac{v - v_o}{v + v_s} \right) = f \left( \frac{v - 0.1v}{v + 0.1v} \right) = f \left( \frac{0.9v}{1.1v} \right) = \frac{9}{11}f \]
• Set up the frequency difference equation:
We are given $f_{\text{app}} - f_{\text{rec}} = 400\text{ Hz}$:
\[ \left( \frac{11}{9} - \frac{9}{11} \right) f = 400 \] \[ \left( \frac{121 - 81}{99} \right) f = 400 \] \[ \frac{40}{99} f = 400 \] \[ f = \frac{400 \times 99}{40} = 10 \times 99 = 990\text{ Hz} \]
Step 4: Final Answer:
The frequency of the source of sound is $990\text{ Hz}$, which corresponds to Option (D).
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