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A block of mass $100 \mathrm{~g}$ is suspended vertically from a massless spring system of spring constant, $k=1 \mathrm{~N} / \mathrm{m}$ 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, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
$4 m$
Given, mass $m=100 \mathrm{~g}=0.1 \mathrm{~kg}$,
spring constant, $k=1 \mathrm{~N} / \mathrm{m}$ and impulse, $I=2 \mathrm{~N} \mathrm{~s}$ 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,
$\begin{array}{rlrl} & & k_p x & =m g+\frac{I}{T / 4} \\ \Rightarrow & & 2 x=0.1 \times 10+\frac{2 \times 4}{1.4} \\ \Rightarrow & & x=3.35 \mathrm{~m}=4 \mathrm{~m}\end{array}$
Hence, the maximum displacement from the equilibrium position of the block is $4 \mathrm{~m}$.
spring constant, $k=1 \mathrm{~N} / \mathrm{m}$ and impulse, $I=2 \mathrm{~N} \mathrm{~s}$ 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,
$\begin{array}{rlrl} & & k_p x & =m g+\frac{I}{T / 4} \\ \Rightarrow & & 2 x=0.1 \times 10+\frac{2 \times 4}{1.4} \\ \Rightarrow & & x=3.35 \mathrm{~m}=4 \mathrm{~m}\end{array}$
Hence, the maximum displacement from the equilibrium position of the block is $4 \mathrm{~m}$.
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