Question

The displacement of a particle executing simple harmonic motion is given by $y =A_{0}+ A\, sin\,wt + B\, cos\,wt$
Then the amplitude of its oscillation is given by:

• A. $\sqrt {A^2_0 +(A+B)^2}$
• B. $A+B$
• C. $A_0 +\sqrt{A^2+B^2}$
• D. $\sqrt{A^2+B^2}$

Answer 1

Answer: (D)

$x=A_{0}+A \sin \,wt+B \,\cos \,w t $
$x-A_{0}=A\, \sin\, wt+B \,\cos \,w t $
$\therefore $ Amplitude $=\sqrt{A^{2}+B^{2}+2 A B \cos 90}$
$=\sqrt{A^{2}+B^{2}}$