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(C) In Simple Harmonic Motion $(SHM)$, the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$, where $\omega$ is the a

Vedclass · Accepted Answer

$3.142$

Vedclass · Answer

(C) In Simple Harmonic Motion $(SHM)$, the acceleration $a$ of a vibrating particle is given by the relation $a = \omega^2 x$, where $\omega$ is the angular frequency and $x$ is the displacement from the mean position.Given: Acceleration $a = 16 \ cm/s^2$ and displacement $x = 4 \ cm$.Substituting these values into the formula:$16 = \omega^2 	imes 4$$\omega^2 = \frac{16}{4} = 4$$\omega = 2 \ rad/s$The time period $T$ is related to angular frequency by the formula $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{2} = \pi \ s$Using $\pi \approx 3.142$, we get $T = 3.142 \ s$.

$A$ particle vibrating simple harmonically has an acceleration of $16 \ cm/s^2$ when it is at a distance of $4 \ cm$ from the mean position. Its time period is: (in $s$)

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