Question

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, $g=10\, m / s ^{2}$ )

• A. 2 m
• B. 4 m
• C. 5 m
• D. 9 m

Answer 1

Answer: (B)

Given, mass $m=100\, g =0 . 1\, kg$,
spring constant, $k=1\, N / m$ and impulse, $I=2 \,N\, s$ 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 =m g+\frac{I}{T / 4}$
$\Rightarrow 2 x =0.1 \times 10+\frac{2 \times 4}{1.4}$
$\Rightarrow x =3.35\, m \simeq 4 m$
Hence, the maximum displacement from the equilibrium position of the block is $4\, m$.