The acceleration of a particle performing S.H | vedclass

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Question

(D) For a particle performing $S.H.M.$,the acceleration $a$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position and $\o

Vedclass · Accepted Answer

$3.14$

Vedclass · Answer

(D) For a particle performing $S.H.M.$,the acceleration $a$ is given by $a = -\omega^2 x$,where $x$ is the displacement from the mean position and $\omega$ is the angular frequency.Given: Displacement $x = 3\;cm$ and acceleration $a = 12\;cm/s^2$.Taking the magnitude: $12 = \omega^2 	imes 3$.$\omega^2 = \frac{12}{3} = 4$.$\omega = 2\;rad/s$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{2} = \pi\;s$.Using $\pi \approx 3.14$,we get $T = 3.14\;s$.

The acceleration of a particle performing $S.H.M.$ at a distance of $3\;cm$ from the mean position is $12\;cm/s^2$. Its time period is ..... $s$.

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