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JEE PHYSICS MCQ: OSCILLATIONS

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Analyzing +4 | -1
The acceleration of a particle moving along xx-axis is a=−200x+50a=-200 x+50. It is released from x=2x=2. Here ' aa ' and ' xx ' are in S.I. units. The motion of particle will be :-
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Correct option: D
Simple harmonic
Option D is correct because the acceleration can be written in the form a = -kx', which indicates simple harmonic motion.

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