Given the equations for the currents: $i_1 = I_{01} \sin(300t)$ $i_2 = I_{02} \sin(300(t + T/6)) = I_{02} \sin(300t + 300T/6)$ Since the angular frequency $\omega = 300 \text{ rad/s}$, we know that $\omega = \frac{2\pi}{T}$, which implies $T = \frac{2\pi}{300} = \frac{\pi}{150} \text{ s}$. (a) Phase difference: The phase of the first current is $\phi_1 = 300t$. The phase of the second current is $\phi_2 = 300t + 300(T/6) = 300t + 50T$. Substituting $T = \frac{\pi}{150}$: $$\Delta \phi = \phi_2 - \phi_1 = 50T = 50 \left( \frac{\pi}{150} \right) = \frac{\pi}{3} \text{ radians}$$ (b) Initial phase of the second current: The initial phase is the value of the phase at $t = 0$: $$\phi_2(0) = 300(0) + 50T = 50T = 50 \left( \frac{\pi}{150} \right) = \frac{\pi}{3} \text{ radians}$$ (c) Frequency of the first current: The angular frequency is $\omega = 300 \text{ rad/s}$. The frequency $f$ is given by: $$\begin{aligned} f &= \frac{\omega}{2\pi} \\ &= \frac{300}{2\pi} \\ &= \frac{150}{\pi} \text{ Hz} \\ &\approx 47.75 \text{ Hz} \end{aligned}$$