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(C) Given the equation of motion: $x(t) = A \sin^2(\alpha t)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$, we can re

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$5$

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(C) Given the equation of motion: $x(t) = A \sin^2(\alpha t)$.Using the trigonometric identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$, we can rewrite the equation as:$x(t) = A \left[ \frac{1 - \cos(2\alpha t)}{2} ight] = \frac{A}{2} - \frac{A}{2} \cos(2\alpha t)$.The general equation for simple harmonic motion $(SHM)$ is given by $x(t) = a_1 + a_2 \cos(\omega t)$, where $\omega$ is the angular frequency.Comparing the given equation with the general form, we identify the angular frequency as $\omega = 2\alpha$.The relationship between angular frequency $\omega$ and time period $T$ is $\omega = \frac{2\pi}{T}$.Given $T = 0.2 \ s$, we have:$2\alpha = \frac{2\pi}{0.2}$$\alpha = \frac{\pi}{0.2} = \frac{\pi}{1/5} = 5\pi \ rad/s$.Therefore, the value of $\alpha$ is $5\pi \ rad/s$.

$A$ particle executes simple harmonic motion according to the equation $x(t) = A \sin^2(\alpha t)$. If the time period of the $SHM$ is $0.2 \ s$, then the value of $\alpha$ (in units of $rad/s$) is (in $pi$)

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