Past exam of the mathematical methods II course of the University of Oxford / 2016 / 11 / b / 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 $\sin (kL)=0$, givingTheir angular frequencies are ${\omega }_n=cn\pi /L$. Unlike the freely translating chain, this fixed string has no translational zero normal mode.