The displacement of a particle executing simple harmonic motion i… - Vidyadip

MCQ

The displacement of a particle executing simple harmonic motion is given by

$\mathrm{y}=\mathrm{A}_{0}+\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$

Then the amplitude of its oscillation is given by

A
$\mathrm{A}_{0}+\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$
✓
$\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$
C
$\sqrt{\mathrm{A}_{0}^{2}+(\mathrm{A}+\mathrm{B})^{2}}$
D
$\mathrm{A}+\mathrm{B}$

✓

Answer

Correct option: B.

$\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$

b
$y=A_{0}+A \sin \omega t+B \cos \omega t$

$y=A_{0}+\sqrt{A^{2}+B^{2}} \sin (\omega t+\phi)$

$\mathrm{A}_{0}$ is mean position, and $\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}$ is amplitude

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