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

Explanation

Given , M = 10 , m = 5 , u = 1 , k = 1
After 12 collisions of mass `M+12 m ` is at rest in its equilibrium position and block of mass m comes with speed u ans collides in - elastically with it . Let velocity of the combined system is v .

`mu = (M+13m ) v `
`v = ("mu")/(M+13m) = u/15`
` v = omega A, "where " omega = sqrt((K)/(M+13m))`
`rArr " " u/15 = sqrt((k)/(M+13m)) xx A rArr A = 1/15 sqrt((75)/1) = 1/(sqrt(3))`