1
NEET 2019
MCQ (Single Correct Answer)
+4
-1
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The displacement of a particle executing simple harmonic motion is given by

y = A0 + A sin$$\omega $$t + B cos$$\omega $$t.

Then the amplitude of its oscillation is given by :
A
$$\sqrt {{A^2} + {B^2}} $$
B
A + B
C
A + $$\sqrt {{A^2} + {B^2}} $$
D
$$\sqrt {A_0^2 + {{\left( {A + B} \right)}^2}} $$