Question

The displacement of the particle varies with time according to the relation. $y=a \sin \omega t+b \cos \omega t, \text { then }$

• A. The motion is oscillating but not SHM
• B. The motion is SHM with amplitude $a + b$
• C. The motion is SHM with amplitude $a^2+b^2$
• D. The motion is SHM with amplitude $\sqrt{a^{2}+b^{2}}$

Answer 1

Answer: (D)

Given, $y=a \sin \omega t+b \cos \omega t$
Let $a=A \cos \theta$ and $b=A \sin \theta$........(i)
then $ y =A \cos \theta \sin \omega t+A \sin \theta \cos \omega t $
$ y=A \sin (\omega t+\theta) $
which is in the form of $SHM$
From Eq. (i)
$ a^{2}+b^{2} =A^{2} \cos ^{2} \theta+A^{2} \sin ^{2} \theta $
$\Rightarrow A=\sqrt{a^{2}+b^{2}} $