A vibrating tuning fork of frequency 512 Hz is held over a water column with one end closed and the other open. As the water level is allowe

Question

A vibrating tuning fork of frequency 512 Hz is held over a water column with one end closed and the other open. As the water level is allowed to fall, a loud sound (resonance) is heard at specific water levels. Assume you start with the tube full of water, and begin steadily lowering the water level. What is the water level (as measured from the top of the tube) for the third such resonance? Take the speed of sound in air to be 343 m/s.

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Neala 2 weeks 2021-07-19T14:32:46+00:00 1 Answers 4 views 0

Answers ( )

    0
    2021-07-19T14:34:13+00:00

    Answer:

    83.74  cm

    Explanation:

     The air column over water level will act as vibrating air column creating standing sound waves . Third resonance occurs when length of water column becomes equal to 5 x λ / 4 . where λ is wave length of sound.

    In resonance frequency of air column and tuning fork becomes equal .

    Here if L be the length of air column for third resonance

    L = 5 x λ / 4

    frequency of tuning fork = 512

    wavelength of sound produced = velocity of sound  / frequency

    343 / 512

    λ = .6699 m

    L = 5 x .6699 / 4

    = .8374 m

    = 83.74  cm .

    water level will be 83.74 cm deep.

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