A particle is executing SHM with time period π sec along a straight line. At t=0, it has started its motion from one of its extreme position. Then the minimum time after which its kinetic energy will be equal to its potential energy is

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$\frac{π}{4} \sec$

$\frac{π}{6} \sec$

$\frac{π}{2} \sec$

$\frac{π}{8} \sec$

answer is D.

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Detailed Solution

Equation of SHM is $x = Acosωt$
At time t , potential energy, $U = \frac{1}{2} {mω}^{2} x^{2} = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt$
And kinetic energy, $K = \frac{1}{2} {mω}^{2} (A^{2} - x^{2}) = \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt$

$When K = U, \frac{1}{2} {mω}^{2} A^{2} \sin^{2} ωt = \frac{1}{2} {mω}^{2} A^{2} \cos^{2} ωt \Rightarrow tanωt = \pm 1$

$\begin{array}{l} \Rightarrow ωt = \frac{π}{4}, \frac{3 π}{4}, \frac{5 π}{4}, \frac{7 π}{4}, . ....... etc \\ ∴ t_{\min} = \frac{π}{4 ω} = \frac{π}{4 \times \frac{2 π}{π}} = \frac{π}{8} \sec \end{array}$