A simple harmonic oscillator has an amplitude | vedclass

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(C) The magnitude of acceleration $a$ for a simple harmonic oscillator at a displacement $x$ from the mean position is given by the formula $a = \omeg

Vedclass · Accepted Answer

$\frac{\pi^2}{4} \ m \ s^{-2}$

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(C) The magnitude of acceleration $a$ for a simple harmonic oscillator at a displacement $x$ from the mean position is given by the formula $a = \omega^2 x$.Here,$\omega$ is the angular frequency,defined as $\omega = \frac{2 \pi}{T}$.Given: Time period $T = 2 \ s$,displacement $x = 0.25 \ m$.First,calculate the angular frequency: $\omega = \frac{2 \pi}{2} = \pi \ rad/s$.Now,substitute the values into the acceleration formula: $a = (\pi)^2 	imes 0.25$.Since $0.25 = \frac{1}{4}$,we get $a = \pi^2 	imes \frac{1}{4} = \frac{\pi^2}{4} \ m \ s^{-2}$.

$A$ simple harmonic oscillator has an amplitude of $0.5 \ m$ and a time period of $2 \ s$. What is the magnitude of acceleration when it is displaced from the mean position by $0.25 \ m$?

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