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(D) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the d

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$3.14$

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(D) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the displacement from the equilibrium position.Given $a = 12 \ cm/sec^2$ and $y = 3 \ cm$.Substituting these values: $12 = \omega^2 	imes 3$.$\omega^2 = \frac{12}{3} = 4$.$\omega = 2 \ rad/sec$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.$T = \frac{2\pi}{2} = \pi \approx 3.14 \ sec$.

If the acceleration of a particle at a distance of $3 \ cm$ from the equilibrium position is $12 \ cm/sec^2$,what is the time period of the simple harmonic motion in seconds?

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