Question

The displacement of the motion of a particle is represent by a equation $y=A \sin \omega t+B \cos \omega t$ The motion of particle is

• A. SHM with amplitude $\sqrt{{A}^{2}+B^{2} }$
• B. SHM with amplitude $A+B$
• C. SHM with amplitude $A$
• D. Oscillatory but not in SHM

Answer 1

Answer: (A)

SHM with amplitude $\sqrt{{A}^{2}+B^{2} }$

Answer 2

The correct answer is A:SHM with amplitude $\sqrt{A^2+B^2}$
$y=A \sin \omega t+B \cos \omega t$
let $A=a \cos \phi$
$B=a \sin \phi$
then equation (1) becomes
$y=a \cos \phi \sin \omega t+a \sin \phi \cos \omega t$
$y=a \sin (\omega t+\phi)$
It is clear that the equation number $(2)$
in simple harmonic motion with amplitude $a$
squaring and adding $(2)$ and $(3)$, we get
$A^{2}+B^{2} \leq a^{2}\left(\cos ^{2} \phi+\sin ^{2} \phi\right)$
$a=\sqrt{A^{2}+B^{2}}$