The Wave Equation — Question 7

Question 7 – Solution

Strategy. A snapshot can hide how many oscillations have occurred; a velocity measurement retains both rate and direction.

Step 1: List all compatible speeds. Since the spatial sine is not identically zero, the snapshot requires $\cos(c\pi\Delta/L)=0$. For positive $c$, $$\boxed{c_n=\frac L\Delta\left(n+\frac 12\right),\qquad n=0,1,2,\ldots.}$$ Every listed speed gives the same zero snapshot, as well as the specified initial displacement and zero velocity. The positive zeros of cosine prove completeness.

Step 2: Use an actual prior restriction. The bound $c<L/\Delta$ allows only $n=0$, so $c=L/(2\Delta)$. Without this independent information, arbitrarily many half-oscillations may have occurred before the snapshot. A zero snapshot does not imply zero speed or determine which crossing of equilibrium was observed.

Step 3: Recover speed from signed velocity. At the midpoint, $$\boxed{v_n=u_t(L/2,\Delta) =(-1)^{n+1}\frac{\pi A}{\Delta}\left(n+\frac 12\right).}$$ A measured $v$ is consistent exactly when $n=|v|\Delta/(\pi A)-1/2$ is a nonnegative integer and its sign is $(-1)^{n+1}$. The strictly increasing magnitudes then select one $n$ uniquely. In particular, a zero velocity measurement is incompatible with the stated nonzero-amplitude zero snapshot.

Step 4: Verify the numerical multiple. For $v=-5\pi A/(2\Delta)$, the magnitude gives $n=2$ and the predicted sign $(-1)^3$ is negative, as measured. Thus $$\boxed{c=\frac{5L}{2\Delta},\qquad u(L/2,\Delta/2)=A\cos(5\pi/4)=-\frac{A}{\sqrt 2}.}$$ The signed result distinguishes this phase from other times at which the displacement vanishes. Inferring the lowest possible speed without a prior bound would produce the wrong prediction for the intermediate snapshot.