Question

A steel wire in a piano has a length of 0.680 m and a mass of 4.600 ✕ 10⁻³ kg. To what tension must this wire be stretched so that the fundamental vibration corresponds to middle C (
`f_C` = 261.6 Hz on the chromatic musical scale)?

Answer 1

728 N

Step-by-step explanation:

`L` = length of the wire = 0.680 m

`m` = mass of the steel wire = 0.0046 kg

`f` = Fundamental frequency = 261.6 Hz

`T` = tension force in the steel wire

Fundamental frequency in wire is given as

`f = (1)/(2L) \sqrt{(TL)/(m) } \\261.6 = (1)/(2(0.680)) \sqrt{(T(0.80))/(0.0046) }\\(2) (0.680) (261.6) = \sqrt{(T(0.80))/(0.0046) }\\355.8 = \sqrt{(T(0.80))/(0.0046) }\\355.8^(2) = (T(0.80))/(0.0046)\\T = ((355.8^(2)(0.0046)))/(0.80) \\T = 728 N`