Oscillations

JEE Main level, numerical answer, medium

34 of 194

Question 34 of 194

A particle of mass 1kg1kg is hanging from a spring of force constant 100Nm−1.100N{m}^{-1}. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T.T. The time when the kinetic energy and potential energy of the system will become equal, is Tn.\dfrac{T}{n}. The value of nn is ________.

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Answer: 8

KE=PEKE=PE
12k(A2−x2)=12kx2\dfrac{1}{2}k\left({A}^{2}-{x}^{2}\right)=\dfrac{1}{2}k{x}^{2}
x2=A22{x}^{2}=\dfrac{{A}^{2}}{2}
x=±A2x=\pm \dfrac{A}{\sqrt{2}}
y=Asin⁡ωty=A\sin \omega t
A2=Asin⁡ωt\dfrac{A}{\sqrt{2}}=A\sin \omega t
sin⁡π4=sin⁡ωt\sin \dfrac{\pi }{4}=\sin \omega t
ωt=π4;t=T8.\omega t=\dfrac{\pi }{4};t=\dfrac{T}{8}. Thus n=8n=8

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