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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. You have a stopped pipe of adjustable length close to a taut 62.0 cm, 7.25 g wire under a tension of 4510 N. You want to adjust the length of the pipe so that, when it produces sound at its fundamental frequency, this sound causes the wire to vibrate in its second overtone with very large amplitude. The speed of sound in air is 344 m/s.

Respuesta :

Answer:

Length of pipe [tex]= 0.057[/tex] meter

Explanation:

Speed of a transverse wave on a string

[tex]v = \sqrt{\frac{F}{\mu} }[/tex]

where F is the tension in string and [tex]\mu[/tex] is the mass per unit length

Thus,

[tex]\mu = \frac{m}{L}[/tex]

Substituting the given values we get -

[tex]\mu = \frac{7.25 * 10^{-3}}{0.62}\\mu = 0.0117 \frac{Kg}{m}[/tex]

Speed of a transverse wave on a string

[tex]v = \sqrt{\frac{4510}{0.0117} } \\v = 620.86 \frac{m}{s}[/tex]

For third harmonic wave , frequency is equal to

[tex]f = \frac{nv}{2L}[/tex]

Substituting the given values, we get -

[tex]f = \frac{3 * 620.86}{2 * 0.62} \\f = 1502.08[/tex]

Length of pipe

[tex]L = \frac{nv}{4 f}[/tex]

Substituting the given values we get

[tex]n = 1[/tex] for first harmonic wave

[tex]L = \frac{344* 1}{4*1502.08} \\L = 0.057[/tex]

Length of pipe [tex]= 0.057[/tex] meter

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