A block of mass m is suspended from one end of a light spring as shown. The origin O is considered at distance equal to natural length of the spring from the ceiling and vertical downwards direction as positive y-axis. When the system is in equilibrium a bullet of mass `(m)/(3)` moving in vertical up wards direction with velocity `v_0` strikes the block and embeds into it. As a result, the block (with bullet embedded into it) moves up and start oscillating. Based on the given information, answer the following question:
Q. The amplitude of oscillation would be

Options

A. `sqrt(((4mg)/(3k))^2+(mv_0^2)/(12k))`
B. `sqrt((mv_0^2)/(12k)+((mg)/(3k))^2)`
C. `sqrt((mv_0^2)/(6k)+((mg)/(k))^2)`
D. `sqrt((mv_0^2)/(6k)+((4mg)/(3k))^2)`

Answer

B

Explanation

Initially in equilibrium let the elongation in spring be `y_0` then `mg=ky_0`
`y_0=(mg)/(k)` As the bullet strikes the block with velocity with velocity `v_0` and gets embed ded into it, the velocity of the combined mass can be computed by ising the principle of moment conservation.
`(m)/(3)v_0=(4m)/(3)vimpliesv=(v_0)/(4)`
Let new mean position is at distance y from the origin, then
`ky=(4m)/(3)gimpliesy=(4mg)/(3k)`
Now, the block executes SHM about mean position defined by `y=(4mg)/(3k)` with time period `T=2pisqrt((4m)/(3k))`. At `t=0`, the combined mass is at a displacement of `(y-y_0)` from mean position and is moving with velocity v, then by using `v=omegasqrt(A^2-x^2)`, we can find the amplitude of motion,
`((v_0)/(4))^2=(3k)/(4m)[A^2-(y-y_0)^2]=(3k)/(4m)[A^2-((mg)/(3k))^2]`
`impliesA=sqrt((mv_0^2)/(12k)+((mg)/(3k))^2)`
To compute the time taken by the combined mass from `y=(mg)/(k)` to `y=0`, we can either go for equation method of circular motion projection method.
Required time, `t=(theta)/(omega)=(alpha-beta)/(omega)`
`cosalpha=(y-y_0)/(A)=(((4mg)/(3k)-(4mg)/(k))/(A))=(mg)/(3kA)`
`cosbeta=(y)/(A)=(4mg)/(3kA)`
So, `t=(cos^-1((mg)/(3kA))-cos^-1((4mg)/(3kA)))/(omega)`
`=sqrt((4m)/(3k))[cos^-1((mg)/(3kA))-cos^-1((4mg)/(3kA))]`