Consider the boundary-value problem introduced in the construction of the mathematical model for the shape of a rotating string:
$$Ty'' + \rho\omega^2y = 0, \quad y(0) = 0, \quad y(L) = 0$$
For constants $T$ and $\rho$, define the critical speeds of angular rotation ωn as the values of $\omega$ for which the boundary-value problem has nontrivial solutions. Find the critical speeds $\omega_n$ and the corresponding deflections $y_n(x)$.
(Give your answers in terms of $n$, making sure that each value of $n$ corresponds to a unique critical speed.)
I set $\rho\omega^2/T = a$, simplifying everything
$a < 0$ and $a = 0$ turn trivial
For $a > 0$ I end up with:
$$c_2\sin(aL) = 0 $$ thus $$ aL = \pi, 2\pi, 3\pi \dots =n\pi $$
$$ a = \frac{n\pi}{L} $$
$$\frac{\rho\omega^2}{T} = \frac{n\pi}{L} $$ leaving me with:
$$\omega_n = \sqrt{\frac{n\pi T}{\rho L}} $$ which unfortunately is wrong..
Where did I go wrong? Any help would be appreciated :)
If $$ a = \frac{\rho\omega^2}{T} $$
You should have
$$ y'' + ay = 0 $$
Then $$ y(x) = \sin (\sqrt{a} x) $$
where
$$ \sqrt{a}L = n\pi $$
so $$ a_n = \left(\frac{n\pi}{L}\right)^2 = \frac{\rho\omega^2}{T} $$ $$ \omega_n = \frac{n\pi}{L}\sqrt{\frac{T}{\rho}}$$