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28 December, 18:41

A glass tube (open at both ends) of length L is positioned near an audio speaker of frequency f = 770 Hz. For what values of L will the tube resonate with the speaker? (Assume that the speed of sound in air is 343 m/s.) m (lowest possible value) m (second lowest possible value) m (third lowest possible value)

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  1. 28 December, 22:00
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    - The lowest possible value of the length L1 = 0.45/2

    L1 = 0.225m

    - Second lowest possible value

    L2 = 0.45m

    - Third lowest possible value

    L3 = 3 (0.45) / 2

    L3 = 0.675m

    Explanation:

    Before calculating the length of the resonance tube, we need to know the wavelength produced by the wave.

    Using the expression

    v = Foλ

    V is the velocity of wave

    Fo is the resonance frequency

    λ is the wavelength

    From the formula

    λ = v/Fo

    λ = 343/770

    λ = 0.45m

    For an open pipe, the first resonant length L1 = λ/2

    Second resonant length L2 = λ

    Third resonant length L3 = 3λ/2

    The lowest possible value of the length L1 = 0.45/2

    L1 = 0.225m

    Second lowest possible value

    L2 = 0.45m

    Third lowest possible value

    L3 = 3 (0.45) / 2

    L3 = 0.675m
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