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JEE Mains · Physics · STD 11 - 13. oscillations

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 ..... .

  1. A \(5\)
  2. B \(6\)
  3. C \(8\)
  4. D \(7\)
Verified Solution

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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