Answer:
T = 708.81 N
Explanation:
Given that,
Length of a steel wire in a piano, l =0.54 m
Mass, [tex]m=4.8\times 10^{-3}\ kg[/tex]
We need to find the tension must this wire be stretched so that the fundamental vibration corresponds to middle C, fc = 261.6 Hz
The equation for fundamental frequency is given by :
[tex]f=\dfrac{1}{2l}\times \sqrt{\dfrac{T}{\mu}} \\\\f=\dfrac{1}{2l}\times \sqrt{\dfrac{T}{\dfrac{m}{l}}} \\\\261.6=\dfrac{1}{2\times 0.54}\times \sqrt{\dfrac{T}{\dfrac{4.8\times 10^{-3}}{0.54}}} \\\\261.6\times 2\times 0.54=\sqrt{\dfrac{T}{\dfrac{4.8\times 10^{-3}}{0.54}}}\\\\282.528=\sqrt{\dfrac{T}{0.00888}} \\\\(282.528)^2=\dfrac{T}{0.00888}\\\\T=708.81\ N[/tex]
So, the required tension in the wire is 708.81 N.
