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A guitar player can change the frequency of a string by "bending" it-pushing it along a fret that is perpendicular to its length. This stretches the string, increasing its tension and its frequency. The B string on a guitar is 64 cm long and has a tension of 74 N. The guitarist pushes this string down against a fret located at the center of the string, which gives it a frequency of 494 Hz. He then bends the string, pushing with a force of 4.0 N so that it moves 8.0 mm along the fret.

* What is the new frequency?

Sagot :

okpalawalter8

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]

 

 

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