A block of mass 2 kg is attached with two identical springs of spring constant 20 N m - 1 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______.

Question

A block of mass 2 kg is attached with two identical springs of spring constant 20 N m - 1 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______.,

MARKS App · Accepted Answer

When the block is displaced by a distance x , force acting on block due to both springs will be in the same direction. Therefore, F = - 2 k x ⇒ a = - 2 k m x , Comparing it with a = - ω 2 x , we get ω = 2 k m = 2 × 20 2 = 20 rad s - 1 Now, T = 2 π ω = 2 π 20 = π 5 Hence, required value of X = 5 .

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