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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 $0.2 \mathrm{~s}$, then the value of $\alpha$ (in units of $\mathrm{rad} / \mathrm{s}$ ) is
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Verified Answer
The correct answer is:
$5 \pi$
Given that, the equation of SHM
$$
\begin{aligned}
x(t) & =A \sin ^2(\alpha t) \\
& =A\left[\frac{1-\cos 2 \alpha t}{2}\right]=\frac{A}{2}-\frac{A}{2} \cos 2 \alpha t
\end{aligned}
$$
We know that, general equation of SHM is
But according to question,
So, comparing Eqs. (i) and (ii), we get
$$
\begin{aligned}
& a_1=\frac{A}{2}, a_2=-\frac{A}{2} \text { and } \omega=2 \alpha \\
& \alpha=\frac{\omega}{2}=\frac{2 \pi}{2 T}=\frac{\pi}{(0.2)} \mathrm{rad} / \mathrm{s} \text { [Given, } T=0.2 \mathrm{~s} \text { ] }
\end{aligned}
$$
Hence, $\quad \alpha=5 \pi \mathrm{rad} / \mathrm{s}$
$$
\begin{aligned}
x(t) & =A \sin ^2(\alpha t) \\
& =A\left[\frac{1-\cos 2 \alpha t}{2}\right]=\frac{A}{2}-\frac{A}{2} \cos 2 \alpha t
\end{aligned}
$$
We know that, general equation of SHM is
But according to question,
So, comparing Eqs. (i) and (ii), we get
$$
\begin{aligned}
& a_1=\frac{A}{2}, a_2=-\frac{A}{2} \text { and } \omega=2 \alpha \\
& \alpha=\frac{\omega}{2}=\frac{2 \pi}{2 T}=\frac{\pi}{(0.2)} \mathrm{rad} / \mathrm{s} \text { [Given, } T=0.2 \mathrm{~s} \text { ] }
\end{aligned}
$$
Hence, $\quad \alpha=5 \pi \mathrm{rad} / \mathrm{s}$
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