Physics · 2019
NEET 2019 · 5 May · Code P1 · Q10
The displacement of a particle executing simple harmonic motion is given by y= A_0+ A sin ω t + B cos ω t. Then the amplitude of its oscillation is…
The displacement of a particle executing simple harmonic motion is given by
$\displaystyle y=\mathrm{A}_0+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$.
Then the amplitude of its oscillation is given by :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \sqrt{\mathrm{A}^2+\mathrm{B}^2}$
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NEET (UG) 2019 Physics question, with the answer from NTA’s final answer key. Where our answers come from.