 Solution

For small transverse displacement, the wave equation on a string iswhere $T$ is the tension and $\rho$ is mass per unit length. Using separation of variables, put $y=X(x)F(t)$, the ratios $X^{\prime \prime }/X=F^{\prime \prime }/(c^{2}F)$ equal a constant. The fixed-end conditions $X(0)=X(L)=0$ exclude nontrivial solutions for nonnegative separation constants; for $-{\kappa }^2$ they give $X=\sin (\kappa x)$ and $\kappa =n\pi /L$, $n\ge 1$. The associated angular frequencies are ${\omega }_n=n\pi c/L$. Thus the general superposition of separated normal modes isThe string is initially straight and at rest before the impulse, so $A_n=0$. Assume the hammer interval lies inside $[0,L]$. The imposed velocity $g(x)=y_t(x,0)$ has Fourier sine series coefficientTherefore $B_n=g_n/{\omega }_n$, giving the amplitude of every excited normal mode.

The string energy is $\frac{\rho }{2}{\int }_0^L(y_t^2+c^2{y}_x^2)dx$. At the impulse it is $E_{\mathrm{total}}=\rho a{v}^2/2$. Orthogonality of the sine modes gives $E_n=\rho L{g}_n^2/4$ initially; each mode subsequently exchanges kinetic and elastic energy while retaining that total. Thus the modal energy of a localized velocity impulse on a string has fractionsFor $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin (n\pi /3)=0$ or $\sin (n\pi /20)=0$. Hence the unexcited modes are precisely