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    Hi can someone please help me solve the following problem:

    The lowest or fundamental frequency for a stringed instrument is given below:

    f 0 = 1/2l √T/m

    f 0 = fundamental or lowest frequency (Hz)
    l = length of string (m)
    T = tension force in string (N)
    m= mass of 1m of the string (kg)

    On a particular piano two adjacent low notes use the same tension in the wire of mass per unit length of 0.030 kg m-1. One of the wires is 0.78m long and it vibrates at a frequency of 55Hz. What is the tension in the string? The other string is slightly longer at 0.91m. What is the frequency of its vibration?
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    (Original post by a5h)
    Hi can someone please help me solve the following problem:

    The lowest or fundamental frequency for a stringed instrument is given below:

    f 0 = 1/2l √T/m

    f 0 = fundamental or lowest frequency (Hz)
    l = length of string (m)
    T = tension force in string (N)
    m= mass of 1m of the string (kg)

    On a particular piano two adjacent low notes use the same tension in the wire of mass per unit length of 0.030 kg m-1. One of the wires is 0.78m long and it vibrates at a frequency of 55Hz. What is the tension in the string? The other string is slightly longer at 0.91m. What is the frequency of its vibration?
    Its a simple question
    f0 = ½l √T/ρ (I prefer using ρ to m since its more of a density than mass)
    f0 = 55Hz = ½ 0.78 √T/0.030 kg m-1
    110 / 0.78 = √T/0.030 kg m-1
    141.026 = √T/0.030 kg m-1
    19888 = T/0.030 kg m-1
    19888 x 0.030 kg m-1 = T
    T = 596.647kg (technically the wrong unit for tension but who gives)

    Then we need to compare f' with f.
    Since f = ½l √T/ρ
    and the only change in the string is l.
    therefore we can reduce the formula to f = l k (where k = ½√T/ρ although this is unimportant)
    Hence we work out k for the first string. 55Hz = 0.78m k
    k = 70.512Hz m-1.
    Then for
    f' = 0.91m k
    f' = 0.91m 70.512Hz m-1
    f' = 64.167Hz
 
 
 
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