Question

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

• A. 2 m
• B. 4 m
• C. 5 m
• D. 9 m

Answer 1

Answer: (B)

Given, mass $m=100g=0.1kg$,
spring constant, $k=1N/m$ and impulse, $I=2Ns$ The given system has parallel connection of the springs.
So, time period, $T=2\pi \sqrt{\frac{m}{k_p}}=2\pi \sqrt{\frac{0.1}{2}}\left(\because k_p=k+k\right)$
$=\frac{2\pi }{\sqrt{20}}=\frac{\pi }{\sqrt{5}}=1.40$
Impulse acts at equilibrium position, so it acts only $\frac{T}{4}$ seconds.
Now, force act on the spring system,
$k_{p}x=mg+\frac{I}{T/4}$
$\Rightarrow 2x=0.1\times 10+\frac{2\times 4}{1.4}$
$\Rightarrow x=3.35m\simeq 4m$
Hence, the maximum displacement from the equilibrium position of the block is $4m$.