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A siren emitting a sound of frequency 80...

A siren emitting a sound of frequency 800 Hz moves away from an observer towards a cliff at a speed of `15ms^-1`. Then the frequency of sound that the observer hears in the echo reflected from the cliff is (Take velocity of sound in air `=330ms^-1`)

A

800Hz

B

838Hz

C

885 Hz

D

765Hz

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the frequency of the sound that the observer hears in the echo reflected from the cliff. We will use the Doppler effect formula for sound. ### Step-by-Step Solution: 1. **Identify the Given Values:** - Frequency of the siren (f) = 800 Hz - Speed of the siren (u_s) = 15 m/s (moving away from the observer) - Speed of sound in air (v) = 330 m/s 2. **Determine the Apparent Frequency Heard by the Cliff:** Since the source (siren) is moving away from the observer, we need to calculate the frequency of the sound when it reaches the cliff. The formula for the apparent frequency (f') when the source is moving away from a stationary observer is: \[ f' = \frac{v}{v + u_s} \cdot f \] Here, \( u_s \) is positive because the source is moving away from the observer. Substituting the values: \[ f' = \frac{330}{330 + 15} \cdot 800 \] \[ f' = \frac{330}{345} \cdot 800 \] \[ f' = 0.9565 \cdot 800 \approx 765.2 \text{ Hz} \] 3. **Determine the Frequency of the Echo Heard by the Observer:** The echo will return to the observer, and now the cliff acts as a source of sound. The observer is stationary, and the cliff is effectively moving towards the observer. The formula for the apparent frequency when the source (cliff) is stationary and the observer is moving towards it is: \[ f'' = \frac{v + u_o}{v} \cdot f' \] Here, \( u_o \) (the speed of the observer) is 0 because the observer is stationary. Substituting the values: \[ f'' = \frac{330 + 0}{330} \cdot 765.2 \] \[ f'' = 1 \cdot 765.2 \approx 765.2 \text{ Hz} \] 4. **Final Calculation for the Echo Frequency:** The frequency of the echo that the observer hears is: \[ f'' = 765.2 \text{ Hz} \] ### Conclusion: The frequency of sound that the observer hears in the echo reflected from the cliff is approximately **765.2 Hz**.
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