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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2021 · 31 August · Morning Shift · Physics · Oscillations

A particle of mass 1kg is hanging from a spring of force constant...

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 ________.

Correct answer

0

Step-by-step solution

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\omegaty = Asin\omegat

A2=Asin\omegat\dfrac{A}{\sqrt{2}} = Asin\omegat
sinπ4=sin\omegatsin\dfrac{\pi}{4} = sin\omegat
\omegat=π4;t=T8.\omegat = \dfrac{\pi}{4}; t = \dfrac{T}{8} . Thus n=8n = 8

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