= Solution
{c}
Substitute the <separated solution> $y=e^{-\alpha t}F(t)\sin(nx)$ into the <damped wave equation>. Its <time derivatives> give
$$
y_{tt}+2\alpha y_t=e^{-\alpha t}(F''-\alpha^2F)\sin(nx),
$$
whereas $y_{xx}=-n^2e^{-\alpha t}F\sin(nx)$. Thus
$$
F''+(n^2-\alpha^2)F=0,\qquad F(0)=0,\quad F'(0)=1.
$$
The last two conditions follow from $y(x,0)=0$ and $y_t(x,0)=\sin(nx)$. For $0<\alpha<n$, let $\nu=\sqrt{n^2-\alpha^2}>0$. The <simple harmonic motion> equation has <solution of a differential equation>
$$
\boxed{F(t)=\frac{\sin(\nu t)}\nu,\qquad \nu=\sqrt{n^2-\alpha^2}.}
$$
This is an <underdamped> <normal mode>: its oscillations have the decaying factor $e^{-\alpha t}$.
Back to article page
