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QJEEM21Ph3f28: JEE Main 2021 Physics - Class 11 JEEM21Ph3f28 Solution: A particle of mass 1 kg is hanging from a spring of force…

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 Tx{T \over x} . The value of x is _____________.

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JEE Main 2021 (Online) 31st August Morning Shift Physics - Simple Harmonic Motion Question 66 English Explanation 1
KE = PE

y=A2=Asin⁡ωty = {A \over {\sqrt 2 }} = A\sin \omega t

JEE Main 2021 (Online) 31st August Morning Shift Physics - Simple Harmonic Motion Question 66 English Explanation 2
t=T8=Txt = {T \over 8} = {T \over x}

x = 8

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