# The potential energy of particle executing simple hormonic motion 0.2 sec after passing the mean position is 1/4 of its total energy. the period of oscillation is

Grade:12

## 1 Answers

Eshan
askIITians Faculty 2095 Points
3 years ago
Dear student,

Potential energy=$\dfrac{1}{2}kx^2$
Total energy=$\dfrac{1}{2}kA^2$
Therefore after t=0.2s, x=A/2
Use$x=Asin \omega t$
Time period$T=\dfrac{2\pi}{\omega}$

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