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A string is stretched betweeb fixed points separated by `75.0 cm`. It observed to have resonant frequencies of `420 Hz` and `315 Hz`. There are no other resonant frequencies between these two. The lowest resonant frequency for this strings isA. `10.5 Hz`B. `105 Hz`C. `1.05 Hz`D. `1050 Hz` |
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Answer» Correct Answer - 2 `(n + 1) (v)/(2l) = 420 …..(1)` `(nv)/(2l) = 315 ….(2)` `(1) - (2) (V)/(mu) = 105 Hz , f_(min) = = 105 Hz` |
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