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NEET UG PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1
The displacement of a particle in SHM various according to the relation x=4(cosπt+sinπt)x = 4\left( {\mathrm{cos\pi }t + \mathrm{sin\pi }t} \right) The amplitude of the particle is
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Correct option: C
424\sqrt 2
Option C (424\sqrt{2}) is correct because the given equation can be rewritten as x=42sin⁡(πt+π4)x = 4\sqrt{2} \sin(\pi t + \frac{\pi}{4}), so the amplitude is 424\sqrt{2}.

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