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Physics · PYQ NEET UG 2019(1)
हिन्दीThe displacement of a particle executing simple harmonic motion is given by \( y = A_0 + A\sin\omega t + B\cos\omega t \). Then the amplitude of its oscillation is given by:
Correct answer · Option D
\( \sqrt{A^2 + B^2} \)
Explanation
The oscillatory part is \( A\sin\omega t + B\cos\omega t \), which can be written as \( R\sin(\omega t + \phi) \) where the amplitude is \( R = \sqrt{A^2 + B^2} \).
