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  4. <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/705ec35d09539e74e6a5a8c951a77dbf-.png /> 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 (π/√x) in SI unit. The value of x is

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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

JEE MainJEE Main 2023Oscillations

Solution:

$ 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$