AP EAMCET · PHYSICS · Oscillations
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
- A \(\mathrm{A}=\mathrm{x}_{\mathrm{o}}, \delta=-\frac{\pi}{6}\)
- B \(\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=-\frac{\pi}{6}\)
- C \(\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{\pi}{6}\)
- D \(\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{5 \pi}{6}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{A}=\mathrm{x}_{\mathrm{o}} \omega^2, \delta=\frac{5 \pi}{6}\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { If } x=x_0 \sin (\omega t-\pi / 6) \\ & \text { Then, } a=-\omega^2 x_0 \sin (\omega t-\pi / 6) \quad\left[\because a=-\omega^2 x\right] \\ & =+\omega^2 x_0 \sin [(\omega t-\pi / 6)+\pi] \\ & =\omega^2 x_0 \sin [\omega t+5 \pi / 6] \end{aligned} \]…
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