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JEE Main 2021 (31 Aug, Shift 1) · Energy in SHM

Energy in SHM question from JEE Main 2021 (31 Aug, Shift 1)

Q64JEE Main 2021 · 31 AUG · SHIFT 1Medium

Energy in SHM

A particle of mass 1 kg is hanging from a spring of force constant 100 Nm ${}^{-1}$ . The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $\frac{T}{x}$ . The value of x is _____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 8

Pulled down and released, so motion starts at the extreme: $x = A\cos\omega t$ .
$K = U$ when $x = \frac{A}{\sqrt{2}} \Rightarrow \cos\omega t = \frac{1}{\sqrt{2}} \Rightarrow \omega t = \frac{\pi}{4}$ .
$t = \frac{\pi/4}{2\pi/T} = \frac{T}{8}$
So $x = 8$ .

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