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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1
Instantaneous acceleration (in ms−2{ms}^{-2} ) of a particle executing S.H.M is given by a=−π4sin⁡(π4t−π6)a=-\frac{\pi}{4} \sin \left(\frac{\pi}{4} t-\frac{\pi}{6}\right). The maximum speed of the particle will occur first time at
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Correct option: D
0.67 s0.67 {~s}
Option D (0.67 {~s}) is correct because maximum speed occurs at the mean position where acceleration is zero, which can be found by setting the acceleration equation to zero and solving for t.

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