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A point mass oscillates along $\mathrm{x}$-axis according to $x=x_0 \sin \left(\omega t-\frac{\pi}{6}\right)$. If the acceleration of the point mass is written as $\mathrm{a}=\mathrm{A} \sin (\omega \mathrm{t}+\delta)$ then
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Verified Answer
The correct answer is:
$\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{5 \pi}{6}$
$$
\begin{aligned}
& \text { If } x=x_0 \sin (\omega t-\pi / 6) \\
& \text { Then, } a=-\omega^2 x_0 \sin (\omega t-\pi / 6) \quad\left[\because a=-\omega^2 x\right] \\
& =+\omega^2 x_0 \sin [(\omega t-\pi / 6)+\pi] \\
& =\omega^2 x_0 \sin [\omega t+5 \pi / 6]
\end{aligned}
$$
So, $A=\omega^2 x_0$ and $\delta=5 \pi / 6$
\begin{aligned}
& \text { If } x=x_0 \sin (\omega t-\pi / 6) \\
& \text { Then, } a=-\omega^2 x_0 \sin (\omega t-\pi / 6) \quad\left[\because a=-\omega^2 x\right] \\
& =+\omega^2 x_0 \sin [(\omega t-\pi / 6)+\pi] \\
& =\omega^2 x_0 \sin [\omega t+5 \pi / 6]
\end{aligned}
$$
So, $A=\omega^2 x_0$ and $\delta=5 \pi / 6$
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