The instantaneous power $P$, measured in watts, in an AC circuit containing an inductance is given by $$ P = vi $$ where $t$ is the time in seconds, $i$ is the current in amperes and $v$ is the voltage in volts. Given that $$v = (120\cos\omega t)V$$ and $$i = (5\sin\omega t)A,$$ where $\omega$ is a constant,
(i) find the period of $P$.
(ii) sketch one cycle of the graph of $P$.
(iii) describe the relationship between $P$ and $P_{1}$ where $$ P_{1} = P\left[ u(t) - u\left( t - \frac{\pi}{2\omega} \right) \right] $$
The answers are
(i) Period of $P = \frac{\pi}{\omega}$ seconds.
(ii)
(iii) $P_{1}$ is a unit step function of $P$. The function of $P$ is switched on at $t = 0$ and off at $t = \frac{\pi}{2\omega}$ seconds.
I'm extremely puzzled, it doesn't make any sense to me. I learnt somewhere that for the graphs $$a\sin(bx-c)$$ and $$a\cos(bx-c),$$ the period is $2\pi/b$. But why is it $\omega/2$? Plus how do you get $300\sin2\omega t$?
We know that $P = vi$ where $v = (120\cos\omega t)$ volts and $i = (5\sin\omega t)$ amperes. Thus we have $$ P = (120\cos\omega t) \times (5\sin\omega t) = 600\sin\omega t\cos\omega t. $$ Now recall that $$ \sin(2x) = 2\sin x\cos x, $$ so we have $$ 600\sin\omega t\cos\omega t = 300\sin(2\omega t). $$ To determine the period of this function, we need to solve $$ \sin(2\omega t_{1}) = \sin(2\omega t_{2}). $$ Remember, the period is how long it takes for the graph to be the same again, which is why we have the above equivalence. The $\sin(x)$ function has a period of $\color{red}{2\pi}$, so $$ \sin(2\omega t_{1}) = \sin(2\omega t_{2}) \iff 2\omega t_{1} - 2\omega t_{2} = \color{red}{2\pi} \iff t_{1} - t_{2} = \frac{2\pi}{2\omega} = \frac{\pi}{\omega}. $$ Thus the period of $300\sin(2\omega t)$ is $\pi/\omega$ seconds.