JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2021
- Hard
- Numerical
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 ..... .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(8\)
Step-by-step Solution
Detailed explanation
\({KE}={PE}\)
\({y}=\frac{{A}}{\sqrt{2}}={A} \sin \omega {t}\)
\({t}=\frac{{T}}{8}=\frac{{T}}{{x}}\)
\(x=8\)
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