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A stretched string of length ` 1 m` fixed at both ends , having a mass of `5 xx 10^(-4) kg` is under a tension of `20 N`. It is plucked at a point situated at `25 cm` from one end . The stretched string would vibrate with a frequency ofA. `400 Hz`B. `100 Hz`C. `200 Hz`D. `256 Hz` |
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Answer» Correct Answer - C At `25 cm` , there will be antinode . So , wire will vibrate in two loops . ` v = (2)/( 2 l) sqrt (( T xx l)/( M)) or v = sqrt((T)/(Ml)) = sqrt((20)/(5 xx 10^(-4) xx 1))` ` = sqrt( 4 xx 10^(4)) Hz = 200 Hz` |
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