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Wave Equation Example

PROBLEM DESCRIPTION
An infinite rope is stretched from x = 0 to x → ∞. The rope’s mass is 0.8(kg)/(m) and is under tension of 100 Newtons. If the beginning of the rope is oscillating in simple harmonic motion with amplitude A at a frequency of ω, find an expression for the solution of the rope. At t = 0, the deflection, u(0, x) = 0.
SOLUTION
Let us begin by establishing an expression for the rope’s head (x = 0) as a function of time. Clearly the sine function fits the simple harmonic motion in the problem description that fits sin(0) = 0.
(1) u0(t) = u(t, x = 0) u0(t) = Asin(ωt)
Now recall from our description of traveling waves that the wave equation can be split as:
(2) (()/(t) − C()/(x))(()/(t) + C()/(x))u = 0
Since we expect a wave to travel down the rope (towards x → ∞), we want a “right” traveling wave and no “left” traveling waves (since there is nothing to the left of the rope’s head and the end of the rope is infinitely far away).
(3) u(x, t) = u0(x − Ct)
Putting this together yields:
(4) u(x, t) = Asin(ωCt − ωx))
The problem is not fully solved yet, though. This solution has an oscillation that continues along the rope towards x → ∞, but recall that the wave equation has disturbances propogating at a constant speed, C. Therefore, we posulate that the solution is:
(5) u(x, t) =  Asin(ωCt − ωx),   if x <  = Ct       0,    if x > Ct
Note that the solution is continuous at x = Ct, which represents the distance that the start of the oscillations has reached. We can test this solution by putting it back into the wave equation itself.
First, examine x < Ct:
(6) (2u)/(t2) − C2(2u)/(t2) = 0  − (ωC)2sin(ωCt − ωx) − C2(ω2sin(ωCt − ωx))? = 0 0 = 0
Thus it is clear from Eqn. 6↑ that our solution (Eqn. 5↑) satisfies the wave equation. We saw that it satisfies the initial conditions and boundary conditions, and so know it is correct.