A point mass oscillates along $\mathrm{x}$-axis according to $x=x_0 \sin \left(\omega…

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
  1. $\mathrm{A}=\mathrm{x}_{\mathrm{o}}, \delta=-\frac{\pi}{6}$
  2. $\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=-\frac{\pi}{6}$
  3. $\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{\pi}{6}$
  4. $\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{5 \pi}{6}$

Solution

$ \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 $\delta=5 \pi / 6$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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