Vibrating String — Question 6

Question 6 – Solution

Strategy. Switch-off is a delayed negative step in forcing; controlling pulse duration can cancel the entire excited state.

Step 1: Superpose the two forcing steps. Only the first spatial mode is driven. Writing $u=q(t)\sin(\pi x)$ gives $q''+\pi^2q=F_0\chi_{(0,\tau)}$ with zero initial data. Hence $$q(t)=\frac{F_0}{\pi^2} \begin{cases} 1-\cos(\pi t),&0\le t\le\tau,\\ \cos(\pi(t-\tau))-\cos(\pi t),&t\ge\tau. \end{cases}$$ The two expressions and their first derivatives agree at $t=\tau$; only acceleration can jump. They verify the forcing on each open time interval.

Step 2: Find exact cancellation durations. For $t\ge\tau$, the cosine difference becomes $$\boxed{q(t)=\frac{2F_0}{\pi^2}\sin\frac{\pi\tau}{2} \sin\bigl(\pi(t-\tau/2)\bigr).}$$ Its displacement and velocity vanish identically precisely when $\sin(\pi\tau/2)=0$, giving $\boxed{\tau=2m,\ m=1,2,\ldots}$. Equivalently, the switch-off state $q(\tau)=F_0(1-\cos\pi\tau)/\pi^2$, $q'(\tau)=F_0\sin\pi\tau/\pi$ is zero exactly for those durations.

Step 3: Evaluate residual energy and work. After switch-off, $$\boxed{E_{\mathrm{res}}=\frac 14(q'^2+\pi^2q^2) =\frac{F_0^2}{\pi^2}\sin^2\frac{\pi\tau}{2}.}$$ During forcing, the power is $\int_0^1F_0\sin(\pi x)u_t\,dx=F_0q'/2$. Its integral from zero to $\tau$ is $F_0q(\tau)/2=F_0^2(1-\cos\pi\tau)/(2\pi^2)$, the same energy.

Step 4: Interpret the pulse-duration curve. The maximum is $F_0^2/\pi^2$, attained for $\tau=2m+1$, $m\ge 0$. For an even-integer duration, positive work during motion in the force direction is canceled by negative work while the string moves against the force. A positive force need not supply positive power at every instant.