Question

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a wire that is 6.93 m long, has a mass of 59.6 g, and is stretched under a tension of 468 N?

Answer 1

Solution)

We know,

F= n(1/2L) ( root(TL/m)

For, lowest frequency, n=1

So, f= 1(1/2*6.93)(root(468*6.93/0.596)

F= 5.322 hz (Ans)

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B) for n=2

We get, F= 10.64 Hz

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C) for n=3

We get, F= 15.96 Hz

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Good luck!:)