A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies f1, f2, f3 and f4(f4>f3>f2>f1) forming 2,3,4 and 5 loops respectively. The ratio of any two resonance frequencies will be minimum for difference in the loops to be:

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answer is C.

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Detailed Solution

Let f1 = mv2L = 2v2L. Similarly f4 = (n+3)2L = 5v2Lf = f1f4 will be minimum f = f1f4 = nn+3 f will be minimum for difference in loops = 3.

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