Physics - SHM Question with Solution | TestHub

Given the equation of motion is

$y=\mathrm{sin\omega }t+\mathrm{cos\omega }t ...\left(1\right)$

Simplify equation (1) to obtain the standard form:

$\begin{array}{rcl}y& =& \sqrt{2}\left(\sin \omega t\cos \frac{\pi }{4}+\cos \omega t\sin \frac{\pi }{4}\right)\\ & =& \sqrt{2}\sin \left(\omega t+\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right) ...\left(2\right)\end{array}$

Comparing equation (2) with standard equation of SHM $y=A\sin \left(\omega t+\varphi \right)$, we get amplitude of the given oscillation as $\sqrt{2}$.