Physics - SHM Question with Solution | TestHub

Given displacement is $x=a\sin \omega t+b\cos \omega t$

Multiplying and dividing by $\sqrt{a^2+b^2}$, we get

$x=\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}\sin \omega t+\frac{b}{\sqrt{a^2+b^2}}\cos \omega t\right)$

Let $\cos \varphi =\frac{a}{\sqrt{a^2+b^2}}$ and $\sin \varphi =\frac{b}{\sqrt{a^2+b^2}}$

$x=\sqrt{a^2+b^2}(\sin \omega t\cos \varphi +\cos \omega t\sin \varphi )$

$x=\sqrt{a^2+b^2}\sin (\omega t+\varphi )$

This represents a simple harmonic motion with amplitude $A=\sqrt{a^2+b^2}$.

Answer: The motion is simple harmonic motion with an amplitude $\sqrt{a^2+b^2}$