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The PE of a simple harmonic oscillator 0.1 s after crossing the mean position is 1/4 of its total energy. Then the period of its oscillation isA. 0.2 sB. 0.3 sC. 0.9 sD. 1.2 s |
Answer» Correct Answer - D `PE=(1)/(4)TE" at x"=(A)/(2)` `x=Asinomegat` `(A)/(2)=Asinomegat therefore sinomegat=(1)/(2)` `therefore omegat=sin^(-1)((1)/(2))=(pi)/(6) therefore (2pi)/(T)t=(pi)/(6)` `therefore T=1.2s` |
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