Vibrating String — Question 1

Question 1 – Solution

Strategy. Initial displacement and initial velocity set different temporal coefficients; spatial orthogonality separates both energy and state recurrence.

Step 1: Normalize the three temporal modes. Fixed ends select $\sin(n\pi x)$ with frequencies $\omega_n=n\pi$. A modal initial velocity $g_n$ gives coefficient $g_n/\omega_n$ multiplying $\sin(\omega_nt)$. Therefore $$\boxed{u=\sin(\pi x)\cos(\pi t) +\tfrac 12\sin(2\pi x)\cos(2\pi t) +\tfrac 23\sin(3\pi x)\sin(3\pi t).}$$ The last coefficient is $2/3$, not $2\pi$.

Step 2: Verify the state and test for a flat snapshot. Each term has equal second derivatives in $x$ and $t$; each spatial sine vanishes at both endpoints. At $t=0$ the first two terms give the stated displacement. Differentiation gives $2\pi\sin(3\pi x)$ as the initial velocity. If $u(\cdot,t)$ were identically zero, orthogonality would force both $\cos(\pi t)=0$ and $\cos(2\pi t)=0$. But the first condition gives $t=k+1/2$, at which $\cos(2\pi t)=-1$. Thus the string is never entirely flat.

Step 3: Add the orthogonal modal energies. For $q_n(t)\sin(n\pi x)$, the energy is $E_n=\tfrac 14[q_n'^2+(n\pi)^2q_n^2]$. Distinct sines are orthogonal in the kinetic term, and distinct cosines in the elastic term, so no cross terms remain. Evaluating at $t=0$ gives $$\boxed{E_1=\frac{\pi^2}{4},\qquad E_2=\frac{\pi^2}{4}, \qquad E_3=\pi^2,\qquad E=\frac{3\pi^2}{2}.}$$ Each modal energy is constant because $q_n''+(n\pi)^2q_n=0$.

Step 4: Find the fundamental state period. All three temporal modes return with their derivatives after time $2$. Conversely, a return of the full state forces its nonzero first mode to satisfy $\cos(\pi P)=1$ and $\sin(\pi P)=0$. Thus $P$ must be a positive even integer. The smallest is $$\boxed{P=2.}$$ A state recurrence requires displacement and velocity together; isolated zeros of one modal displacement do not establish a period.