JEE MAIN - Physics (2023 - 24th January Morning Shift - No. 20)
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A block of a mass 2 kg is attached with two identical springs of spring constant 20 N/m each. The block is placed on a frictionless surface and the ends of the springs are attached to rigid supports (see figure). When the mass is displaced from its equilibrium position, it executes a simple harmonic motion. The time period of oscillation is $$\frac{\pi}{\sqrt x}$$ in SI unit. The value of $$x$$ is ____________.
Answer
5
Explanation
Both the springs are in parallel so net spring constant is $K_{\text {net }}=K_{1}+K_{2}=40 \mathrm{~N} / \mathrm{m}$
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
$$ \begin{aligned} & =2 \pi \sqrt{\frac{2}{40}} \\\\ & =\frac{\pi}{\sqrt{5}} \end{aligned} $$
$\therefore x=5$
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
$$ \begin{aligned} & =2 \pi \sqrt{\frac{2}{40}} \\\\ & =\frac{\pi}{\sqrt{5}} \end{aligned} $$
$\therefore x=5$
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