InterviewSolution
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InterviewSolution
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Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops, and 4 loops, the frequency are in the ratio 1:2:3:4. |
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Answer» Length for each loop = \(\frac{\lambda}{2}\) Now, L = \(\frac{n\lambda}{2}\) λ = \(\frac{2L}{n}\) (1) But v = vλ or λ = \(\frac{v}{u}\) Putting in eqn, (1) \(\frac{v}{u}=\frac{2L}{n}\) v = \(\frac{n}{2L}\)u v = \(\frac{n}{2L}\sqrt{\frac{T}{\mu}}\) [∵ u = \(\sqrt{\frac{T}{\mu}}\) ] For n = 1, v1 = \(\frac{1}{2L}\sqrt{\frac{T}{m}}\) = v0 For n = 2, v2 = \(\frac{1}{2L}{\sqrt{\frac{T}{m}}}\) = 2v0 Therefore, u1: u2:u3: u4 = n1: n2: n3:n4 u1: u2:u3: u4 = 1: 2: 3: 4 |
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