Answer:
728 N
Explanation:
[tex]L[/tex] = length of the wire = 0.680 m
[tex]m[/tex] = mass of the steel wire = 0.0046 kg
[tex]f[/tex] = Fundamental frequency = 261.6 Hz
[tex]T[/tex] = tension force in the steel wire
Fundamental frequency in wire is given as
[tex]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[/tex]
