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Q.

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

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a

1 : 4

b

3 : 1

c

2 : 1

d

4 : 1

answer is C.

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Detailed Solution

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$\begin{array}{l} when \\ t = \frac{T}{12}, then x = A sin \frac{2 π}{T} \times \frac{T}{12} = \frac{A}{2} \\ KE = \frac{1}{2} {mv}^{2} = \frac{1}{2} {mω}^{2} (A^{2} - x^{2}) \\ = \frac{1}{2} {mω}^{2} (A^{2} - \frac{A^{2}}{4}) \\ = \frac{3}{4} (\frac{1}{2} {mω}^{2} A^{2}) \\ PE = \frac{1}{2} {mω}^{2} x^{2} = \frac{1}{4} (\frac{1}{2} {mω}^{2} A^{2}) \\ \frac{KE}{PE} = \frac{3}{1} \end{array}$

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