Physics - SHM Question with Solution | TestHub

(a) $(\cos \omega t+\sin \omega t)$ is a periodic function. It can also be written as $$\begin{array}{rl}& =\frac{\sqrt{2}}{\sqrt{2}}\sin \omega t+\frac{\sqrt{2}}{\sqrt{2}}\cos \omega t\\ & =\sqrt{2}\left(\cos \frac{\pi }{4}\sin \omega t+\sin \frac{\pi }{4}\cos \omega t\right)\\ & =\sqrt{2}\sin \left(\omega t+\frac{\pi }{4}\right)=\sqrt{2}\sin \left(\omega t+\frac{\pi }{4}+2\pi \right)\\ & =\sqrt{2}\sin \left[\omega \left(t+\frac{2\pi }{\omega }\right)+\frac{\pi }{4}\right]\end{array}$$ This represent a simple harmonic function with period $\frac{2\pi }{\omega }$ and phase $\frac{\pi }{4}$. be written as $$\begin{array}{rl}& =\sqrt{2}\left[\sin \omega t\cos \frac{\pi }{4}-\cos \omega t\sin \frac{\pi }{4}\right]\\ & =\sqrt{2}\sin \left(\omega t-\frac{\pi }{4}\right)=\sqrt{2}\sin \left[\omega \left(t+\frac{2\pi }{\omega }\right)-\frac{\pi }{4}\right]\end{array}$$ This represent a simple harmonic function with period $\frac{2\pi }{\omega }$. period $\frac{2\pi }{\omega }$. (c) $F(t)=1-\sin 2\omega t$ This is a non periodic function. (d) $F(t)=\sin \omega t+\cos (\omega t+\alpha )$ also represent a simple harmonic function.