= Solution
Transverse force balance on a short element gives
$$
\rho y_{tt}=\tau y_{xx},\qquad c=\sqrt{\tau/\rho}.
$$
With fixed ends,
$$
y(x,t)=\sum_{n\ge1}[A_n\cos(\omega_nt)+B_n\sin(\omega_nt)]\sin\frac{n\pi x}{L},
\quad\omega_n=\frac{n\pi c}{L}.
$$
Here $A_n=0$. If $b_n$ are the sine coefficients of initial <velocity>, then
$$
b_n=\frac{4v}{n\pi\sqrt\varepsilon}\sin\frac{n\pi l}{L}\sin\frac{n\pi\varepsilon}{2L},
\qquad B_n=\frac{b_n}{\omega_n}.
$$
The total deposited energy is $E=\rho v^2/2$, while
$$
\frac{E_n}{E}=\frac{8L}{n^2\pi^2\varepsilon}
\sin^2\frac{n\pi l}{L}\sin^2\frac{n\pi\varepsilon}{2L}.
$$
The seventh mode is eliminated by striking at a node $l=jL/7$, $j=1,\ldots,6$. For a narrow hammer, each fixed-mode fraction is $O(\varepsilon)$; beyond the inverse-width cutoff the envelope falls as $n^{-2}$, modulated by the two sine factors.
Back to article page
