For a particle performing S.H.M.,the equation | vedclass

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(D) The correct option is $D$.Concept: The standard differential equation for $S.H.M.$ is given by $\frac{d^2 x}{dt^2} + \omega^2 x = 0$,where $\omega

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$\frac{2 \pi}{\sqrt{\alpha}}$

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(D) The correct option is $D$.Concept: The standard differential equation for $S.H.M.$ is given by $\frac{d^2 x}{dt^2} + \omega^2 x = 0$,where $\omega$ is the angular frequency.Comparing the given equation $\frac{d^2 x}{dt^2} + \alpha x = 0$ with the standard equation,we get $\omega^2 = \alpha$,which implies $\omega = \sqrt{\alpha}$.The time period $T$ of the motion is defined as $T = \frac{2 \pi}{\omega}$.Substituting the value of $\omega$,we get $T = \frac{2 \pi}{\sqrt{\alpha}}$.

For a particle performing $S.H.M.$,the equation is $\frac{d^2 x}{dt^2} + \alpha x = 0$. Then the time period of the motion will be

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