InterviewSolution
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InterviewSolution
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A string is stretched between fixed points separated by `75.0cm`. It is observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string isA. (a) `105 Hz`B. (b) `1.05 Hz`C. ( c ) `1050 Hz`D. (d) `10.5 Hz` |
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Answer» Correct Answer - A (a) Given `(nv)/(2l) = 315` and `(n + 1)(v)/(2l) = 420` rArr `(n + 1)/(n) = (420)/(315)` rArr `n = 3` Hence `3 xx (v)/(2l) = 315` rArr `(v)/(2l) = 105 Hz` lowest resonant frequency is when `n = 1` Therefore lowest resonant frequency `= 105 Hz`. |
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