= Solution
For small transverse displacement, the <wave equation on a string> is
$$
\boxed{y_{tt}=c^2y_{xx},\qquad c^2=T/\rho,}
$$
where $T$ is the tension and $\rho$ is mass per unit length. Using <separation of variables>, put $y=X(x)F(t)$, the ratios $X''/X=F''/(c^2F)$ 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 $\kappa=n\pi/L$, $n\ge1$. The associated angular frequencies are $\omega_n=n\pi c/L$. Thus the general superposition of separated <normal modes> is
$$
\boxed{y(x,t)=\sum_{n\ge1}\left[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)\right]\sin\frac{n\pi x}{L}.}
$$
The 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> coefficient
$$
g_n=\frac{2v}{L}\int_{l-a/2}^{l+a/2}\sin\frac{n\pi x}{L}\,dx
=\frac{4v}{n\pi}\sin\frac{n\pi l}{L}\sin\frac{n\pi a}{2L}.
$$
Therefore $B_n=g_n/\omega_n$, giving the amplitude of every excited <normal mode>.
The string energy is $\frac\rho2\int_0^L(y_t^2+c^2y_x^2)\,dx$. At the impulse it is $E_{\mathrm{total}}=\rho av^2/2$. Orthogonality of the sine modes gives $E_n=\rho Lg_n^2/4$ 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 fractions
$$
\boxed{\frac{E_n}{E_{\mathrm{total}}}=\frac{8L}{a\pi^2n^2}\sin^2\frac{n\pi l}{L}\sin^2\frac{n\pi a}{2L}.}
$$
For $l=L/3$ and $a=L/10$, a mode is absent exactly when $\sin(n\pi/3)=0$ or $\sin(n\pi/20)=0$. Hence \b[the unexcited modes are precisely]
$$
\boxed{n\in3\mathbb N_{>0}\ \cup\ 20\mathbb N_{>0}.}
$$
Back to article page
