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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

9493488568 , 6 Years ago
Grade 12
anser 1 Answers
Eshan

Last Activity: 6 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
Usex=Asin \omega t
Time periodT=\dfrac{2\pi}{\omega}

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