Vibrating String — Question 4

Question 4

A string of length $L$ has tension $T>0$, density $\mu>0$ and speed $c=\sqrt{T/\mu}$. Its left end is fixed and its right end is free for $t>0$: $$u_{tt}=c^2u_{xx},\qquad u(0,t)=0,\quad u_x(L,t)=0.$$ At release its displacement is $f(x)=hx/L$, $h>0$, and its velocity is zero. Interpret this as a finite-energy problem, since the initial slope need not satisfy the newly released end condition.

Tasks

1. Derive the mixed-end eigenfunctions, frequencies and coefficients of $f$. Construct the solution.

2. Explain the initial corner incompatibility. Prove that the full state reverses sign after $2L/c$ and has smallest positive period $4L/c$.

3. Show that at $t=L/c$ the string is flat but is not at rest. Find its velocity in the open interval and verify the conserved energy.

4. For $L=c=h=1$, sketch exact profiles at $t=0,1/2,1$. Use the odd reflection at the fixed end and even reflection at the free end, and mark the moving corner.

Question 4 – Solution

Strategy. A free end selects half-integer modes; the release launches a corner rather than making the initial slope compatible retroactively.

Step 1: Construct the mixed-mode solution. The conditions $X(0)=0$, $X'(L)=0$ give $k_n=(2n+1)\pi/(2L)$, $n\ge 0$, and $\omega_n=ck_n$. Since $\int_0^L\sin^2(k_nx)\,dx=L/2$, $$b_n=\frac 2L\int_0^L\frac{hx}{L}\sin(k_nx)\,dx =\frac{8h(-1)^n}{(2n+1)^2\pi^2},\qquad \boxed{u=\sum_{n\ge 0}b_n\sin(k_nx)\cos(\omega_nt).}$$

Step 2: State the trace and recurrence precisely. Initially $f'(L)=h/L\ne 0$, so a solution with continuous $u_x$ at the release corner cannot satisfy the new free-end condition there. The displayed finite-energy solution recovers $f$ uniformly and zero initial velocity in $L^2$; its boundary condition has the usual weak/natural meaning at wavefront events. Every active frequency is an odd multiple of $\pi c/(2L)$. Thus both $u$ and $u_t$ change sign after $2L/c$ and return after $4L/c$. The nonzero lowest mode requires that full period, proving minimality.

Step 3: Resolve the flat but moving state. At $t=L/c$ every temporal cosine vanishes. For $0<x<L$, $$u_t(x,L/c)=-\frac{4hc}{L\pi} \sum_{n\ge 0}\frac{\sin((2n+1)\pi x/(2L))}{2n+1} =\boxed{-hc/L}.$$ Here $\sum_{n\ge 0}\sin((2n+1)z)/(2n+1)=\pi/4$ for $0<z<\pi$, the sine series of a constant on that interval. The fixed-end velocity remains zero; the interior velocity trace is an $L^2$ statement, not corner continuity. Initially $E=Th^2/(2L)$. At this flat snapshot, the kinetic energy is $\mu L(hc/L)^2/2=Th^2/(2L)$, so the energy is unchanged.

Step 4: Follow the released-end corner. Extend $f$ oddly across zero and evenly across $L$, with period $4L$, and take the average of its two translates. For the stated units and $0\le t\le 1$ this gives $$\boxed{u(x,t)=\min(x,1-t),\qquad 0\le x\le 1.}$$ For $0<t<1$, the corner is at $x=1-t$: the left part retains slope one and the right part is flat and moving downward. The snapshots below use exact line segments.