NEET-XII-Physics
P1 year:2019
- #10The displacement of a particle executing
simple harmonic motion is given by
``y = A_0 + Asinωt + Bcosωt``
Then the amplitude of its oscillation is given
by :
(1) ``A_0 + \sqrt{A^2 + B^2}``
(2) ``\sqrt{A^2 + B^2}``
(3) `` \sqrt{ A_0^2 +(A + B)^2}``
(4) A + BdigAnsr: 2Ans : ( 2 )
Sol. B
y = ``A_0 + Asinωt + Bsinωt``
Equate SHM
``y' = y - A_0 = Asinωt + Bcosωt``
Resultant amplitude
``R= \sqrt{(A^2+B^2 + 2ABcos90)}``
= ``\sqrt{A^2 + B^2}``