A block of 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 $..........$
JEE MAIN 2023, Medium
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$F =-2 kx , a =-\frac{2 kx }{ m }, \omega=\sqrt{\frac{2 k }{ m }}=\sqrt{\frac{2 \times 20}{2}}$
$=\sqrt{20}\,rad / s$
$T =\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{20}}=\frac{\pi}{\sqrt{5}}$
$x =5$
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