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

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______.

TestHub · 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,

$F = - 2 k x \Rightarrow a = - \frac{2 k}{m} x,$

Comparing it with $a = - ω^{2} x$ , we get

$ω = \sqrt{\frac{2 k}{m}} = \sqrt{\frac{2 \times 20}{2}}$

$= \sqrt{20} rad s^{- 1}$

Now, $T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{20}} = \frac{π}{\sqrt{5}}$

Hence, required value of $X = 5$ .

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