Physics - SHM Question with Solution | TestHub

Given equation of motion is $x=A\sin \omega t+B\cos \omega t$.

Let $A=R\cos \varphi$ and $B=R\sin \varphi$.

Squaring and adding, we get $A^2+B^2=R^2({\cos }^2\varphi +{\sin }^2\varphi )=R^2$.

Therefore, $R=(A^2+B^2{)}^{\frac{1}{2}}$.

Substituting $A$ and $B$ in the given equation:

$x=R\cos \varphi \sin \omega t+R\sin \varphi \cos \omega t$

$x=R\sin (\omega t+\varphi )$

This is the standard equation of simple harmonic motion.

Thus, the motion is simple harmonic with amplitude $R=(A^2+B^2{)}^{\frac{1}{2}}$.

Answer: simple harmonic with amplitude ${\left(A^2+B^2\right)}^{\frac{1}{2}}$