Oscillations

JEE Main level, numerical answer, medium

116 of 194

Question 116 of 194

A block of mass 2kg2kg is attached with two identical springs of spring constant 20Nm−120N{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 πX\dfrac{\pi }{\sqrt{X}} in SI unit. The value of XX is______.

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Answer: 5

When the block is displaced by a distance xx, force acting on block due to both springs will be in the same direction. Therefore,
F=−2kx⇒a=−2kmx,F=-2kx \Rightarrow a=-\dfrac{2k}{m}x,
Comparing it witha=−ω2xa=-{\omega }^{2}x , we get
ω=2km=2×202\omega =\sqrt{\dfrac{2k}{m}}=\sqrt{\dfrac{2\times 20}{2}}
=20rads−1=\sqrt{20}rad{s}^{-1}
Now, T=2πω=2π20=π5T=\dfrac{2\pi }{\omega }=\dfrac{2\pi }{\sqrt{20}}=\dfrac{\pi }{\sqrt{5}}
Hence, required value of X=5X=5.

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