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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1
A particle is subjected to three SHMsSHMs

X1=3sin⁡ωtX_1=3 \sin \omega t

X2=4cos⁡ωtX_2=4 \cos \omega t

X3=6sin⁡(ωt+π)X_3=6 \sin (\omega t+\pi)

Where XX is in metre. The resultant amplitude of particle is
View answer and explanation
Correct option: D
5  m5\, {~m}
Option D (5  m5 \, {\text{~m}}) is correct because it is the resultant amplitude calculated by vector addition of the SHMs.

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