AC circuit: Frequency of the generator

Question:
There is a AC circuit with a resistor $R=50 \ \Omega$. When the time is zero, the voltage is equal to zero. After 1/720 s, the voltage becomes a half of the maximum
voltage. Find the frequency of the voltage generator.

Answer:
The voltage varies with time, so
\[
V = V_m \sin \omega t
\]
where $V_m$ is the peak voltage. From the condition, we can express
\[
0.5V_m = V_m \sin \left(\omega\cdot \frac{1}{720}\right)
\]
$V_m$ is cancelled out. Then take the arcsine.
\[
 \frac{\omega}{720} = \sin^{-1}0.5
\]
Therefore,
\[
\omega = 120 \pi \ \mathrm{rad/s}
\]
In order to get the frequency, divide it by $2\pi$.
\[
f = \frac{\omega}{2\pi} = 60 \ \mathrm{Hz}
\]
Note that we don't have to use the resistance.

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