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19 April, 22:00

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. What is the frequency of its fundamental mode of vibration?

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  1. 19 April, 23:22
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    408.25 Hz.

    Explanation:

    The fundamental frequency of a stretched string is given as

    f' = 1/2L√ (T/m') ... Equation 1

    Note: The a steel piano wire is a string

    Where f' = fundamental frequency of the wire, L = length of the wire, T = tension on the wire, m' = mass per unit length of the wire.

    Given: L = 0.4 m, T = 800 N,

    Also,

    m' = m/L where m = mass of the steel wire = 3.00 g = 3/1000 = 0.003 kg.

    L = 0.4 m

    m' = 0.003/0.4 = 0.0075 kg/m.

    Substituting into equation 1

    f' = 1 / (2*0.4) [√ (800/0.0075) ]

    f' = 1/0.8[√ (106666.67) ]

    f' = 326.599/0.8

    f' = 408.25 Hz.

    Hence the frequency of the fundamental mode of vibration = 408.25 Hz.
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