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)
$a^2+b^2=A^2{\cos }^2\theta +A^2{\sin }^2\theta$
$\Rightarrow A=\sqrt{a^2+b^2}$