Vibrating String — Question 3

Question 3 – Solution

Strategy. A velocity profile must be integrated in time; its jumps do not permit a globally classical interpretation of the initial state.

Step 1: Convert the strike to temporal sine amplitudes. The velocity coefficients are $$g_n=2V\int_{1/3}^{2/3}\sin(n\pi x)\,dx =\frac{2V}{n\pi}\left(\cos\frac{n\pi}{3}-\cos\frac{2n\pi}{3}\right).$$ Because the initial displacement is zero, $$\boxed{u(x,t)=\sum_{n\ge 1}\frac{g_n}{n\pi} \sin(n\pi t)\sin(n\pi x).}$$ The additional division by $n\pi$ is essential to recover $u_t(\cdot,0)=g$.

Step 2: Interpret the symmetry and traces. The strike satisfies $g(1-x)=g(x)$, so only the symmetric, odd-indexed sine modes can occur. The displayed coefficient vanishes for every even $n$ and is nonzero for every odd $n$: the cosine difference for odd $n$ is either $1$ or $-2$. Displacement converges uniformly since its coefficients are $O(n^{-2})$. Velocity recovers $g$ in $L^2$; its initial sine series takes the average $V/2$ at either jump. The PDE holds weakly everywhere and classically away from the reflected characteristic fronts.

Step 3: Compute the midpoint before reflected arrivals. For $0\le t\le 1/2$, the interval $[1/2-t,1/2+t]$ remains within $[0,1]$. The velocity-only d’Alembert formula therefore gives $$u(\tfrac 12,t)=\frac 12\int_{1/2-t}^{1/2+t}g(s)\,ds =\boxed{\begin{cases}Vt,&0\le t\le 1/6,\\V/6,&1/6\le t\le 1/2.\end{cases}}$$ At $t=1/6$, both edges of the moving integration interval pass the strike edges. The one-sided midpoint velocities change from $V$ to zero. A zero subsequent velocity does not undo the displacement already accumulated.

Step 4: Match physical and modal energies. Initially $u_x=0$, so $$\boxed{E=\frac 12\int_0^1g^2\,dx=\frac{V^2}{6}.}$$ Parseval with $\int_0^1\sin^2(n\pi x)\,dx=1/2$ gives $\int g^2=\tfrac 12\sum g_n^2$, hence $E=\tfrac 14\sum g_n^2$. Equivalently, $$\sum_{n\ge 1}\frac{\left[\cos(n\pi/3)-\cos(2n\pi/3)\right]^2}{n^2} =\frac{\pi^2}{6}.$$ Each mode trades kinetic and elastic energy while preserving its contribution $g_n^2/4$.