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)

In the string mass system shown in the figure, the string is compresse

1.	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)
Answer» `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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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)

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