Vibrating String — Question 9

Question 9 – Solution

Strategy. A prescribed end motion launches a ramp front; reflection reverses its displacement sign and transfers the injected energy into a growing slope.

Step 1: Launch the incoming ramp. For $0\le t\le L/c$, $$u_{\mathrm{in}}(x,t)=V\bigl(t-(L-x)/c\bigr)_+.$$ It vanishes ahead of the front $x=L-ct$ and equals $Vt$ at $x=L$. Every function of $t+x/c$ satisfies the wave equation distributionally; here it is classical away from the front. At $t=0$, displacement is zero and velocity is zero at every interior point.

Step 2: Reflect at the fixed left end. The reflected ramp must cancel the incoming displacement at $x=0$: $$\boxed{u=V\bigl(t-(L-x)/c\bigr)_+ -V\bigl(t-(L+x)/c\bigr)_+.}$$ The left trace is identically zero. At $x=L$, the second term is zero through $t=2L/c$, giving $u(L,t)=Vt$. After that time it would change the right trace, so additional reflections would be needed. The formula is not asserted for later times.

Step 3: Balance energy and injected power. Before the first reflection, the moving portion has length $ct$, with $u_t=V$ and $u_x=V/c$. Since $T/c^2=\mu$, its energy is $E=\mu cV^2t$. For $L/c<t<2L/c$, the reflected left portion has length $ct-L$, velocity zero and slope $2V/c$; the right portion has length $2L-ct$, velocity $V$ and slope $V/c$. Hence again $$E=2\mu V^2(ct-L)+\mu V^2(2L-ct) =\boxed{\mu cV^2t}.$$ For $0<t<2L/c$, the right-end slope is $V/c$, so the input power is $T u_x(L,t)u_t(L,t)=TV^2/c=\mu cV^2=E'$.

Step 4: Interpret the corner and the final snapshot. The initial interior velocity is zero, but the driven endpoint velocity jumps to $V$: no continuous velocity at that release corner is possible. Displacement stays continuous across each front, while first derivatives jump; the weak wave equation remains valid. At $t=2L/c$, $u=2Vx/c$ and the open-interval velocity is zero; energy is entirely elastic, $2\mu LV^2$. The endpoint still moves, so this is not a smooth state of rest that persists afterward.