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Repeat the above problem aussuming B is ...

Repeat the above problem aussuming `B` is placed on `A` at a distance `(a)/(2)` from mean position.

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`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))`
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