In the figure shown mass `2m` is at rest and in equilibrium. A particle of mass `m` is released from height `(4.5mg)/(k)` from palte. The particle sticks to the plate. Neglecting the duration of collision. Starting form the when the particles sticks to plate to the time when the spring is in maximum compression for the first time is `2pisqrt((m)/(ak))` then find `a`.

Answer

5

Explanation

Velocity of the particle just before collision, `u=sqrt(2gxx(4.5m g)/(K)) Rightarrow u=3gsqrt((m)/(K))`
Now it collides with the plate.
Now just after collision velocity (V) of the system of ‘plate + particle’
`m u =3mVRightarrow(u)/(3)=gsqrt((m)/(K))`
Now the system performs SHM with time period `T=2pisqrt(3m//K)(omega=sqrt((K)/(3m)))`and mean position as `mg//K` distance below the point of collision.
Let the equation of motion is `y=Asin(omegat+phi)`....(i)
`v=(dy)/(dt)=Asinomegacos(omegat+phi)`....(ii) At t=0,`y=(mg)/(K)` and `v=gsqrt((m)/(K))`
From Eqs.(i)and(ii),`(mg)/(K)=sinphi`....(iii)
`A=(2mg)/(K)`......(iv)
From Eqs. (iii)and (iv), `phi=(5pi)/(6)`, `y=(2mg)/(K)sin(sqrt((k)/(3m))t+(5pi)/(6))`
Hence equation of SHM should be `y=-A=-(2mg)/(K)=(2mg)/(K)sin (sqrt((k)/(3m))t+(5pi)/(6))`
The plate will be at rest again when `y=-A=-(2mg)/(K)=(2mg)/(K)sin (sqrt((k)/(3m))t+(5pi)/(6))`
`Rightarrow sin(sqrt((k)/(3m))t+(5pi)/(6))=-1=sin(3pi)/(2)Rightarrow sqrt((K)/(3m))t+(5pi)/(6)=(3pi)/(2)Rightarrow t=(2pi)/(3)sqrt((3m)/(K))`
Using values , `t=pi//5s`.