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

Answer:

728 N

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

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