JEE Mains · Physics · STD 11 - 13. oscillations
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 \(..........\)

- A \(5\)
- B \(4\)
- C \(3\)
- D \(2\)
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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