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A 1.65-m-long wire having a mass of 0.100 kg is fixed at both ends. The tension in the wire is maintained at 16.0 N. (a) What are the frequencies of the first three allowed modes of vibration

Sagot :

Answer:

Explanation:

mass per unit length ρ = .100 / 1.65 = .0606 . kg /m

length of wire L = 1.65 m

For fundamental frequency , the expression is as follows

n = [tex]\frac{1}{2L} \sqrt{\frac{T}{m} }[/tex]

L = 1.65 , T = 16 n and m = .0606

n = [tex]\frac{1}{2\times 1.65} \sqrt{\frac{16}{.0606} }[/tex]

= 4.9 /s .

This is fundamental frequency .

other mode of vibration ( first three ) will be as follows

4.9 x 2 = 9.8 /s ,

4.9 x 3 = 14.7 /s .

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