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AveGali [126]
3 years ago
10

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 th

at the fundamental vibration corresponds to middle C (f_C = 261.6 Hz on the chromatic musical scale)?
Physics
1 answer:
garik1379 [7]3 years ago
3 0

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

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see answer below

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<h2>B = 3.63x10⁻⁸ T</h2><h2></h2>

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