The 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 : [NEET 2019]

{A}_{0}+\sqrt{ {A}^{2}+ {B}^{2}}

NEET 2019 Physics and Chemistry

Question : 74 of 90

Marks: +1, -0

The displacement of a particle executing simple harmonic motion is given by y=A0+A‌sin‌ω‌t+B‌cos‌ω‌t
Then the amplitude of its oscillation is given by :

[NEET 2019]

√A2+B2
√A02+(A+B)2
A + B
A0+√A2+B2

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