Given:
$J_c = \sigma \sin(\omega t)$
$J_d = \omega~ \varepsilon \cos(\omega t)$
What is the frequency $\omega$ at which $J_c = J_d$?
Here's my work:
$\sigma \sin(\omega t) = \omega~ \varepsilon \cos(\omega t)$
$\sigma = \omega~\epsilon$
$\boxed{\omega = \frac{\sigma}{\varepsilon}}$
Is this formula for $\omega$ valid?
It seems to me that the phase wouldn't match if we equate only the the amplitudes of the two sinusoids, since the left-hand-side (Sin) is out of phase with the right-hand-side (cos)?
However, the physics book i'm looking at says its valid... I just wanted to understand why this is true.
Hint. You want to solve an equation of the form $$A\cos\omega t+B\sin \omega t=0.$$ You can always write $\text{LHS}$ as $R\sin(\omega t+\phi),$ where $R$ and $\phi$ depend on $A$ and $B.$ Then you simply want to solve something of the form $\sin y=0,$ whose solutions are $y=πk$ for any integer $k.$