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 \(..........\)

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Answer & Solution

Correct Answer

(A) \(5\)

Step-by-step Solution

Detailed explanation

\(F =-2 kx , a =-\frac{2 kx }{ m }, \omega=\sqrt{\frac{2 k }{ m }}=\sqrt{\frac{2 \times 20}{2}}\)

\(=\sqrt{20}\,rad / s\)

\(T =\frac{2 \pi}{\omega}=\frac{2 \pi}{\sqrt{20}}=\frac{\pi}{\sqrt{5}}\)

\(x =5\)

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