InterviewSolution
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InterviewSolution
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A wire is stretched between two rigid supports. It is observed that wire resonates at the frequencies `f_1,f_2,f_3` and `f_4(f_4gtf_3,f_2gtf_1)` 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:A. `1`B. `2`C. `3`D. zero |
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Answer» Correct Answer - C Let `f_1=(mv)/(2L)=(2v)/(2L)`. Similarly `f_4=((n+3)v)/(2L)=(5v)/(2L)` `f=(f_1)/(f_4)` will be minimum `f=(f_1)/(f_4)=(n)/(n+3)` `f` will be minimum for difference in loops `=3`. |
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