WBJEE · Physics · Oscillations
The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass \(m\) is represented by \(y=2 \sin \left(\frac{\pi t}{2}+\phi\right) \mathrm{cm}\). The maximum acceleration of the particle is
- A \(\frac{\pi}{2} \mathrm{~cm} / \mathrm{sec}^2\)
- B \(\frac{\pi}{2 \mathrm{~m}} \mathrm{~cm} / \mathrm{sec}^2\)
- C \(\frac{\pi^2}{2 \mathrm{~m}} \mathrm{~cm} / \mathrm{sec}^2\)
- D \(\frac{\pi^2}{2} \mathrm{~cm} / \mathrm{sec}^2\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi^2}{2} \mathrm{~cm} / \mathrm{sec}^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given } y=2 \sin \left(\frac{\pi \mathrm{t}}{2}+\phi\right) \mathrm{cm} \\ & \mathrm{a}_{\max }=\omega^2 \mathrm{~A} \\ & =\left(\frac{\pi}{2}\right)^2 \cdot 2=\frac{\pi^2}{2} \mathrm{~cm} / \mathrm{sec}^2\end{aligned}\)
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