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In Melde,s experiment,the string vibrates in 6 loops when a 30 g weight is placed inthepan of weight 10 g. To make the string to vibrates in 8 loops the10 g. mass that has to be removed form the pan is |
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Answer» ``20 g by Melde's method. The frequency of vibration of a string of length `l,` MASS per unit length m and vibrating in p loops under tension T is given by `n=p/(2l)sqrt(T/m) rArr psqrt(T) =` constant if n, `l andm ` are constant. HENCE, `T prop 1/p^(2) rArrT_(1)/T_(2) = p_(2)^(2)/p_(1)^(2)` or `((30+10))/T_(2) = (8)^(2)/(6)^(2) rArr 40/T_(2) = 64/36` `therefore T_(2)=34/64xx 40 =22.5` So, WEIGHT removed form the pan `=40 - 22.5` `=17.59` |
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