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A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received).
    Correct answer is '6'. Can you explain this answer?
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    A stationary source emits sound of frequency f0 = 492 Hz. The sound is...
    Frequency of sound as received by large car approaching the source.

    This car now acts as source for reflected sound wave

    frequency of sound received by source,


    = 6 Hz
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    A stationary source emits sound of frequency f0 = 492 Hz. The sound is...
    To solve this problem, we can use the Doppler effect equation:

    f' = (v + vd)/(v + vs) * f

    Where:
    - f' is the observed frequency of the sound (after reflection)
    - v is the speed of sound in air (approximately 343 m/s)
    - vd is the velocity of the car towards the source (approaching velocity)
    - vs is the velocity of the source (which is stationary)

    Given:
    - f0 = 492 Hz (frequency emitted by the source)
    - vd = 2 m/s (velocity of the car towards the source)
    - vs = 0 m/s (stationary source)

    Plugging in the given values into the Doppler effect equation, we can calculate the observed frequency (f').

    f' = (343 + 2)/(343 + 0) * 492
    f' = 345/343 * 492
    f' ≈ 496.2 Hz

    Therefore, the observed frequency of the sound after reflection is approximately 496.2 Hz.
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    A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received).Correct answer is '6'. Can you explain this answer?
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    A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received).Correct answer is '6'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received).Correct answer is '6'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz ? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received).Correct answer is '6'. Can you explain this answer?.
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