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JEE PHYSICS MCQ: OSCILLATIONS

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Displacement-time equation of a particle executing SHM is,

x = 4\sin \omega t + 3\sin \,\,(\omega t + \pi /3)

Here

x

is in centimeter and t in second. The amplitude of oscillation of the particle is approximately

View answer and explanation

Correct option: B

6\mathrm{ }cm

Option B (

6\mathrm{ }cm

) is correct because the amplitude of the resulting SHM is found by vector addition of the individual SHMs.

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