A block with mass `M` attached to a horizontal spring with force constant `k` is moving with simple harmonic motion having amplitude `A_(1)`. At the instant when the block passes through its equilibrium position a lump of putty with mass `m` is dropped vertically on the block from a very small height and sticks to it.
(a) Find the new amplitude and period.
(b) Repeat part (a) for the case in which the putty is dropped on the block when it is at one end of its path.

Explanation

(i) Before the lump of putty is dropped the total mechanical energy of the block and spring is
`E_(1)=(1)/(2)kA_(1)^(2)`
Since, the block is at the equilibrium position , U=0, and the energy is purely kinetic . Let `v_(1)` be the speed of the block at the equilibrium position , we have `E_(1)=(1)/(2)Mv_(1)^(2)=(1)/(2)kA_(1)^(2)impliesv_(1)sqrt((k)/(M))A_(1)`
During the process momentum of the system in horizontal direction is conserved. Let `v_(2)` be the speed of the combined mass, then
`(M+m)v_(2)=Mv_(1)`
`therfore v_(2)=(M)/(M+m)v_(1)`
Now, let `A_(2)` be the amplitude afterwards. Then,
`E_(2)=(1)/(2)kA_(2)^(2)=(1)/(2)(M+m)v_(2)^(2)`
Subsituting the proper values, we have
`A_(2)=A_(1)sqrt((M)/(M+m))`