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JEE PHYSICS MCQ: OSCILLATIONS

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The amplitude of a simple harmonic oscillator is 2 cm. At certain instant its PE and KE are equal. The displacement from mean position at that instant is
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Correct option: A
2 cm\sqrt 2 \,\mathrm{cm}
Option A (2 cm\sqrt 2 \,\mathrm{cm}) is correct because at the point where potential and kinetic energies are equal, the displacement is A/2\sqrt{2}, where A is the amplitude.

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