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21 July, 12:52

A string is stretched to a length of 308 cm and both ends are fixed. If the density of the string is 0.023 g/cm, and its tension is 967 N, what is the fundamental frequency? Answer in units of Hz

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  1. 21 July, 13:00
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    Frquency=3,994Hz

    Explanation:

    Tension = 967N

    Density of string (μ) = 0.023g/cm

    Length of the stretched spring=308cm

    Fundamental frequency for nth harmonic:

    Fn=n/2L (√T/μ)

    Substituting the given values to find the frequency:

    f1=1/2 (308cm) * (0.01m/1cm) [ (√967N) / (0.023g/cm) (0.1kg) / (0.1kg/m) / (1g/cm) ]

    =6.16m[ (√967N) / 0.0023kg/m) ]

    =3,994.20Hz

    Approximately,

    The frequency will be = 3,994Hz
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