A block of mass 100 g is suspended vertically from a massless spring system of spring constant, k=1 N/m each. The block is hit from above to impart an impulse of 2 Ns. Calculate the maximum displacement from the Equilibrium position of the block. (Take, `g=10m//s^(2)`)

Options

A. 2m
B. 4 m
C. 5m
D. 9 m

Answer

B

Explanation

Given mass m=100g =0.1 kg,
spring constant, k=1 N/m and impulse, I=2Ns The given ssystem has parallel connection of the springs.
So, time period. `T=2pisqrt((m)/(k_(p)))=2pisqrt((0.1)/(2))("":'k_(p)=k+k)`
`=(2pi)/(sqrt20)=(pi)/(sqrt5)=1.40`
Impulse acts at equilibrium position, so it acts only `(T)/(4)` seconds.
Now force act on the spring system,
`k_(p)x=mg+(I)/(T//4)`
`implies2x=0.1xx10+(2xx4)/(1.4)`
`impliesx=3.35m=4m`
Hence, the maximum displacement from the equilibrium postion of the block is 4 m.