= Solution
The fixed-end <boundary conditions> require $f(0)=f(L)=0$. For $\alpha=0$, the spatial solution is linear and both endpoints force it to vanish; for $\alpha>0$, the exponential solutions likewise cannot satisfy both conditions nontrivially. Write $\alpha=-k^2$ with $k>0$. Then $f=A\sin(kx)$ and $\sin(kL)=0$, giving
$$
\boxed{\alpha_n=-\left(\frac{n\pi}{L}\right)^2,\qquad f_n(x)=A\sin\left(\frac{n\pi x}{L}\right),\qquad n=1,2,\ldots.}
$$
Their <angular frequencies> are $\omega_n=cn\pi/L$. Unlike the freely translating chain, this fixed string has no translational zero <normal mode>.
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