The air pressure p(x, t) in an organ pipe is governed by the wave equation
∂2p / ∂x2 = 1 / c2 ∂2p / ∂t2, 0 < x < l, 0 < t,
where l is the length of the pipe, and c is a physical constant. If the pipe is open, the boundary conditions are given by
p(0, t) = p0 and p(l, t) = p0.
If the pipe is closed at the end where x = l, the boundary conditions are
p(0, t) = p0 and ∂p / ∂x (l, t) = 0.
Assume that c = 1, l = 1, and the initial conditions are
p(x, 0) = p0 cos 2πx, and ∂p / ∂t (x, 0) = 0, 0 ≤ x ≤ 1.
a. Approximate the pressure for an open pipe with p0 = 0.9 at x = 1/2 for t = 0.5 and t = 1, using Algorithm 12.4 with h = k = 0.1.
b. Modify Algorithm 12.4 for the closed-pipe problem with p0 = 0.9, and approximate p(0.5, 0.5) and p(0.5, 1) using h = k = 0.1.