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A particle starts oscillating simple harmonically from its equilibrium position with time period T. The ratio of KE and PE of the particle at time $t = \frac{T}{12}$ is

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$1 : 4$

$2 : 1$

$3 : 1$

$4 : 1$

answer is C.

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Detailed Solution

$Let x = Asin (ωt)$

When

$t = \frac{T}{12}, then x = A \sin \frac{2 π}{T} \times \frac{T}{12} = \frac{A}{2}$

$KE = \frac{1}{2} {mω}^{2} (r^{2} - x^{2}) = \frac{1}{2} {mω}^{2} (A^{2} - \frac{A^{2}}{4}) = \frac{3}{4} (\frac{1}{2} {mω}^{2} A^{2})$

$PE = \frac{1}{2} {mω}^{2} (x^{2}) = \frac{1}{4} (\frac{1}{2} {mω}^{2} A^{2})$

Thus, $\frac{KE}{PE} = \frac{3}{1}$

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