Question 9.3: Find the transverse displacement of a vibrating string of le...
Find the transverse displacement of a vibrating string of length l with fixed endpoints if the string is initially in its rest position and has a velocity distribution g(x).
We look for the solution of the boundary value problem
\frac{∂^{2}y}{∂t^{2}}= c^{2} \frac{∂^{2}y}{∂x^{2}} , (9.5)
where y = y(x, t), with
y(0, t) = 0, y (l, t) = 0,y(x, 0) = 0, \frac{∂}{∂t}y(x, t)|_{t=0}= g(x).
We use the separation ansatz y = X(x) · T (t). By inserting it into (9.5), one obtains
X · \ddot{T} = c^{2}X^{″}T or \frac{X^{″}}{X}(x) =\frac{\ddot{T}}{c^{2}T}(t ). (9.7)
Since the left-hand side of (9.7) depends only on x, the right side only on t , and x and t are independent of each other, the equation is satisfied only then if both sides are constant. The constant is denoted by −λ^{2}.
\frac{X^{″}}{X} =−λ^{2} and \frac{\ddot{T}}{c^{2}T} =−λ^{2},or, transformed,
X^{″} +λ^{2}X = 0 and \ddot{T} + λ^{2}c^{2}T = 0. (9.8)
The two equations have the solutions
X = A_{1} cos λx +B_{1} sin λx, T = A_{2} cos λct +B_{2} sin λct .Since y = X · T , we have
y(x, t) = (A_{1} cos λx + B_{1} sin λx)(A_{2} cos λct +B_{2} sin λct). (9.9)
From the condition y(0, t) = 0, it follows that A_{1}(A_{2} cos λct + B_{2} sin λct) = 0. This condition is satisfied by A_{1} = 0. Then
y(x, t) = B_{1} sin λx(A_{2} cos λct + B_{2} sin λct).We now set
B_{1}A_{2} = a, B_{1}B_{2} = b,and it follows that
y(x, t) = sin λx(a cos λct +b sin λct). (9.10)
From the condition y(l, t) = 0, it follows that sin λl = 0. This happens if
λ l = nπ or λ = \frac{nπ}{l}. (9.11)
Here, n = 1, 2, 3, . . .. The value n = 0 which seems possible at first sight leads to y(x, t) ≡ 0 and must be excluded. The relation (9.11) is inserted into (9.10). The normal vibration will be labeled by the index n:
y_{n}(x, t) = sin \frac{nπx}{l} \left(a_{n} cos \frac{nπct}{l} +b_{n} sin \frac{nπct}{l}\right). (9.12)
Because y(x, 0) = 0, all a_{n} = 0, we have
y_{n}(x, t) = b_{n} sin \frac{nπx}{l} sin \frac{nπct}{l}. (9.13)
By differentiation of (9.13), we get
\frac{∂y_{n}}{∂t} = b_{n} \frac{nπc}{l} sin \frac{nπx}{l} cos \frac{nπct}{l}. (9.14)
For linear differential equations, the superposition principle holds, so that the entire solution looks as follows:
\frac{∂y}{∂t}=\sum\limits_{n=1}^{∞}{\frac{nπcb_{n}}{l} sin \frac{nπx}{l} cos \frac{nπct}{l}}. (9.15)
Because
\frac{∂}{∂t}y(x, t) |_{t=0}= g(x),it follows that
g(x) =\sum\limits_{n=1}^{∞}{\frac{nπcb_{n}}{l} sin \frac{nπx}{l}}. (9.16)
The Fourier coefficients then follow by
\frac{nπcb_{n}}{l}= \frac{2}{l} \int\limits_{0}^{l}{g(x) sin \frac{nπx}{l}dx}or
b_{n} = \frac{2}{nπc} \int\limits_{0}^{l}{g(x) sin \frac{nπx}{l}dx}By inserting (9.18) into (9.13), we obtain the final solution for y(x, t):
y(x, t) = \sum\limits_{n=1}^{∞}{\left(\frac{2}{nπc} \int\limits_{0}^{l}{g(x^{′}) sin \frac{nπx^{′}}{l} dx^{′}}\right) sin \frac{nπx}{l} sin \frac{nπct}{l}}. (9.19)