For a particle executing SHM, determine the r | vedclass

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(B) The acceleration of a particle in $SHM$ is given by $a(t) = -\omega^2 x = -\omega^2 A \cos(\omega t)$.At the extreme position, $x = A$ $(t = 0)$,

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$\frac{2}{\pi}$

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(B) The acceleration of a particle in $SHM$ is given by $a(t) = -\omega^2 x = -\omega^2 A \cos(\omega t)$.At the extreme position, $x = A$ $(t = 0)$, and at the equilibrium position, $x = 0$ $(t = \frac{\pi}{2\omega})$.The average acceleration $A_{	ext{avg}}$ is defined as $\frac{1}{\Delta t} \int_{0}^{\Delta t} a(t) dt$.Here, $\Delta t = \frac{\pi}{2\omega}$.$A_{	ext{avg}} = \frac{1}{\pi / 2\omega} \int_{0}^{\pi / 2\omega} \omega^2 A \cos(\omega t) dt = \frac{2\omega}{\pi} [A \sin(\omega t)]_{0}^{\pi / 2\omega} = \frac{2\omega A}{\pi} (1 - 0) = \frac{2\omega^2 A}{\pi}$.Since the maximum acceleration $A_{\max} = \omega^2 A$, the ratio is $\frac{A_{	ext{avg}}}{A_{\max}} = \frac{2\omega^2 A / \pi}{\omega^2 A} = \frac{2}{\pi}$.

For a particle executing $SHM$, determine the ratio of the average acceleration of the particle between the extreme position and the equilibrium position with respect to the maximum acceleration.

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