An oscillator of mass M is at rest in its equilibrium position in a potential `V=(1)/(2)k(x-X)^(2)`. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. The process repeats every time the oscillator crosses its equilibrium position. The amplitude of oscilllations after 13 collisions is : (M=10, m=5, u=1 ,k=1)

Options

A. `(1)/(2)`
B. `(1)/(sqrt(3))`
C. `(2)/(3)`
D. `sqrt((3)/(5))`

Answer

B

Explanation

In first collision mu momentum will be imparted to system, in second collision when momentum of `(M+m)` is in opposite direction mu momentum of particle will make its momentum zero.

`"mu"=(M+13m)vimpliesv=("mu")/(M+13m)=(u)/(15)`
`v=omega Aimplies(u)/(15)=sqrt((K)/(M-13m))xxA`
Putting value of M, m, u and K we get amplitude
`A=(1)/(15)sqrt((75)/(1))=(1)/(sqrt(3))`