## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

(a) $\phi_0 = \frac{5\pi}{3}$ (b) The phase at point 1 is $\frac{4\pi}{3}$ The phase at point 2 is $0$ The phase at point 3 is $\frac{2\pi}{3}$
(a) The general equation for the motion of the particle is: $x(t) = A~cos(\omega~t+\phi_0)$ We can find $\phi_0$ when $x(0) = \frac{A}{2}$ $cos(\phi_0) = \frac{1}{2}$ $\phi_0 = arccos(\frac{1}{2})$ $\phi_0 = \frac{\pi}{3}, \frac{5\pi}{3}$ In the graph, we can see that a standard cos-curve has been shifted to the left by an angle of $\frac{5\pi}{3}$. Therefore, the phase constant is $\phi_0 = \frac{5\pi}{3}$. (b) We can find the phase at point 1: $phase = \pi+\frac{\pi}{3} = \frac{4\pi}{3}$ We can find the phase at point 2: $phase = 0$ We can find the phase at point 3: $phase = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$