At particle is executing `SHM` with amplitude `A` and has maximum velocity `v_(0)`. Find its speed when it is located at distance of `(A)/(2)` from mean position.

Answer

`(sqrt(3)v_(0))/(2)`

Explanation

`v_(0) = Aomega rArr omega = (v_(0))/(A)`
`v^(2) = omega^(2) (A^(2) - x^(2))`
`v^(2) = ((v_(0))/(A))^(2) (A^(2) - (A^(2))/(2^(2)))`
`v^(2) = v_(0)^(2) (3)/(4), v = (sqrt(3t))/(2)v_(0)`