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.

Explanation

`T = 2pisqrt((m)/(K))`
`T = 2pisqrt((2m)/(K))`

By conservation of momentum
`"mu" = 2mv rArr v = (u)/(2)`
Kinetic Energies at `(a)/(2)`
For mass `m : (1)/(2) "mu"^(2) = (1)/(2) momega^(2)[a - ((a)/(2))^(2)].....(1)`
For mass `2m : (1)/(2) 2mv^(2) = (1)/(2) 2m((omega)/(sqrt(2)))^(2)[A - ((a)/(2))^(2)]......(2)`
`(1)/(2)2m(u^(2))/(4) = (1)/(2)2m(omega^(2))/(2)[A^(2) - (a^(2))/(4)].....(3)`
Dividing equation `(1)` & `(3)`
`2 = (a^(2) - (a^(2))/(4))/(A^(2) - (a^(2))/(4))`
New Amplitude `A = sqrt((5)/(8))a` Ans. `T = 2pisqrt((2m)/(K)`, Amplitude `= asqrt((5)/(8))`