A particle executes simple harmonic motion ac | vedclass

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(C) The given equation is $4 \frac{d^2 x}{d t^2} + 320 x = 0$.Dividing by $4$,we get $\frac{d^2 x}{d t^2} + 80 x = 0$.Since the acceleration $a = \fra

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$\frac{\pi}{2 \sqrt{5}} \ s$

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(C) The given equation is $4 \frac{d^2 x}{d t^2} + 320 x = 0$.Dividing by $4$,we get $\frac{d^2 x}{d t^2} + 80 x = 0$.Since the acceleration $a = \frac{d^2 x}{d t^2}$,we can write $a = -80 x$.Comparing this with the standard $S.H.M.$ equation $a = -\omega^2 x$,we get $\omega^2 = 80$.Therefore,$\omega = \sqrt{80} = \sqrt{16 	imes 5} = 4 \sqrt{5} \ rad/s$.The time period $T$ is given by $T = \frac{2 \pi}{\omega}$.Substituting the value of $\omega$,we get $T = \frac{2 \pi}{4 \sqrt{5}} = \frac{\pi}{2 \sqrt{5}} \ s$.

$A$ particle executes simple harmonic motion according to the equation $4 \frac{d^2 x}{d t^2} + 320 x = 0$. Its time period of oscillation is .........

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