Vibrating String — Question 7

Question 7 – Solution

Strategy. Time orthogonality separates frequencies, but it cannot restore a spatial mode whose node lies at the sensor.

Step 1: Project the temporal record. On $[0,2]$, distinct integer-frequency sines and cosines are orthogonal and each squared norm is one. Consequently $$\boxed{f_n=\frac{\int_0^2h_s(t)\cos(n\pi t)\,dt}{\sin(n\pi s)},\qquad g_n=\frac{n\pi\int_0^2h_s(t)\sin(n\pi t)\,dt}{\sin(n\pi s)}}$$ whenever $\sin(n\pi s)\ne 0$. These recover both initial displacement and velocity coefficients.

Step 2: Identify rational-position blindness. For $s=p/q$ in lowest terms, $\sin(n\pi s)=0$ exactly when $q$ divides $n$. Every such mode contributes zero to $h_s(t)$ for every time and for every choice of its two coefficients. Thus even a perfect continuous trace cannot distinguish states that differ only in those modes.

Step 3: Combine the two sensors. The midpoint misses the even modes; the one-third sensor misses multiples of three. Their common invisible modes are exactly the multiples of six. For $N\le 5$, each mode has a nonzero factor at at least one sensor, so the formulas recover the entire state. For $N\ge 6$, arbitrary sixth-mode displacement or velocity can be added without changing either record. The largest fully identifiable cutoff is therefore $$\boxed{N=5.}$$

Step 4: Distinguish visibility from stability. If $s$ is irrational, no $ns$ is an integer, so each denominator is nonzero. But the supplied approximation gives $|\sin(n\pi s)|=|\sin(\pi(ns-k))|\le\pi/Q$. As $Q\to\infty$, the selected indices cannot stay in a finite set: each such index has a fixed positive distance from the integers. Hence arbitrarily large modes have arbitrarily small sensor factors. An error $e_n$ in the cosine projection becomes $e_n/\sin(n\pi s)$ in $f_n$; the sine projection has the additional factor $n\pi$ for $g_n$. Exact visibility at all modes does not provide a uniform noise bound.