In the string mass system shown in the figure, the string is compressed by `x_(0) = (mg)/(2k)` from its natural length and block is relased from rest. Find the speed o the block when it passes through P (`mg//4k` distance from mean position)
Text Solution
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`omega = sqrt((3k)/(m)), x = Asin(omegat + phi), v = Aomegacos(omegat+phi)` at `t = 0, x = 0 rArr phi = 0` `x = A sinomegat rArr (mg)/(4k) = (mg)/(2k)sin omegat rArr sinomegat = 1/2 rArr omegat = (pi)/(6) rArr t = (pi)/(6omega) = T/12` `v = Aomegacosomegat. V = (mg)/(2k)sqrt((3k)/(m)) cos[sqrt((3k)/(m)) (2pisqrt((m)/(3k))//12)] = gsqrt((9m)/(16k))`
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