Find the time period ($T$) for an electric field wave: $E=E_0\sin{m t}\sin{2mt}$

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Find the time period ($T$) for an electric field wave: $E=E_0\sin{m t}\sin{2mt}$

I thought $T$ is such that, $E(T+t) = E(t)$. As period of given sinusoidal function $E$ is 2$\pi$, $$ \Rightarrow 2\pi = mT \Rightarrow T = 2\pi/m$$

But, I was wrong. How to do it correctly?


The correct answer is:

$6π/m$


Edit: I plotted the graph after suggestion by PM 2Ring on fooplot and got the period as $2\pi$ (in graph, $m =1$).

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5
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You have correctly stated the condition for periodicity is $E(t+T)=E(t)$ but then proceed to ignore that, in particular, the form of E.

Using this and the hint given by PM2Ring, I’m sure you can work it out.

0
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The period is indeed $2\pi/m$. What may appear confusing is that the field is not a harmonic function of time, but rather a sum of two harmonic functions, as it has been noted in the comments: $$E(t) = E_0\sin mt \sin 2mt = \frac{E_0}{2}\left[\cos m t - \cos 3 m t\right].$$ The period of $\cos m t$ is $T_1=2\pi/m$, whereas the period of $\cos 3 m t$ is $T_2=2\pi/(3m)$, i.e. three times shorter. Since the two periods are commensurate (i.e. their ratio is an integer, $T_1/T_2=3$), the whole wave is still periodic and repeats itself after $T=T_1 = 2\pi/m$! So this is the correct answer, although the person who posed the question probably expected the above cited reasoning about the two periodic waves.

Note for moderators
This may appear as a complete answer to a homework question. Note however that this homework question is badly formulated and I needed to provide all the details in order to clarify it.