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Answer:
[tex]f'=504hz[/tex]
Explanation:
From the question we are told that
The B string on a guitar is 64 cm long
The B string tension tension of 74 N.
Frequency of 494 Hz
Pushed with a Force of 4.0 N
It moves 8.0 mm along the fret.
Generally the equation for frequency of ring under tension is mathematically given as
[tex]2Lf=\sqrt{x\frac{T}{\mu} }[/tex]
[tex]2*(64/100)*494=\sqrt{\frac{74}{m/0.64}[/tex]
[tex](632.32)^2={\frac{74}{m/0.64}[/tex]
[tex](632.32)^2=74*{\frac{0.64}{m}[/tex]
[tex](632.32)^2={\frac{47.36}{m}[/tex]
[tex]m=1.18450761*10^-^4[/tex]
Therefore finding the New frequency f'
[tex]f'=\frac{(\sqrt{\frac{74+11}{(\frac{1.18450761*10^-^4}{0.642})}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{74+11}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{85}{1})*(\frac{0.642}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{(\sqrt{(\frac{54.57}{1.18450761*10^-^4}})}{2*0.642}[/tex]
[tex]f'=\frac{678.7471973}{2*0.642}[/tex]
[tex]f'=572.9111096hz[/tex]