Physics - SHM Question with Solution | TestHub

The phase of a particle in S.H.M. at any instant $t$ is given by $\varphi =\omega t+{\varphi }_0=\frac{2\pi }{T}t+{\varphi }_0$

Given $T=16\text{ s}$

Phase at $t=2\text{ s}$ is ${\varphi }_1=\frac{2\pi \times 2}{16}+{\varphi }_0=\frac{\pi }{4}+{\varphi }_0$

Phase at $t=4\text{ s}$ is ${\varphi }_2=\frac{2\pi \times 4}{16}+{\varphi }_0=\frac{\pi }{2}+{\varphi }_0$

Phase difference $\Delta \varphi ={\varphi }_2-{\varphi }_1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$

Answer: $\frac{\pi }{4}$