Frequency of a solution to a PDE

Question

Why is the frequency of $u=\sum_{n=1}^{\infty}B_n$sin$(n\pi x/L)$cos($n^2 \pi^2ct/L)$ equal to the coefficent of $t$ with $L>0,c$ constants, i.e

Why is it proportional to $n^2$?

Answer 1

Quite simply, the instantaneous radian frequency of a harmonic vibration is the time derivative of the argument of the cosine or sine (or complex exponential) describing the vibration. This is the definition for what we mean by frequency. In your case, the time dependence for the $n^\text{th}$ mode is

$$ T_n(t) = \cos(\phi_n(t)) $$ with $$ \phi_n(t)=n^2\pi^2 ct/L $$ Now we apply the definition of instantaneous radian frequency: $$ \omega_n(t) = \phi_n'(t) = n^2\pi^2c/L $$

This is why the frequency is the coefficient that goes with $t$; it's defined as $\frac{d}{dt}$ of the argument and that operation gives the coefficient in front of any terms linear in $t$. Finally, as to why it is proportional to $n^2$ (instead of $n$ like the normal wave equation): this is not the wave equation, so why would you expect it to have the same properties as it? The extra derivatives in space (compared to the wave equation) causes this "extra" factor of $n$ to appear in the frequency when compared to the wavenumber.