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).

Question

Accepted Answer

We look for the solution of the boundary value problem

[latex]\frac{∂^{2}y}{∂t^{2}}= c^{2} \frac{∂^{2}y}{∂x^{2}} ,[/latex] (9.5)

where [latex]y = y(x, t)[/latex], with

[latex]y(0, t) = 0, y (l, t) = 0,[/latex]

[latex]y(x, 0) = 0, \frac{∂}{∂t}y(x, t)|_{t=0}= g(x).[/latex]

We use the separation [latex]ansatz y = X(x)[/latex] · T (t). By inserting it into (9.5), one obtains

[latex]X · \ddot{T} = c^{2}X^{″}T or \frac{X^{″}}{X}(x) =\frac{\ddot{T}}{c^{2}T}(t ).[/latex] (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 [latex]−λ^{2}[/latex].

[latex]\frac{X^{″}}{X} =−λ^{2} and \frac{\ddot{T}}{c^{2}T} =−λ^{2},[/latex]

or, transformed,

[latex]X^{″} +λ^{2}X = 0 and \ddot{T} + λ^{2}c^{2}T = 0.[/latex] (9.8)

The two equations have the solutions

[latex]X = A_{1} cos λx +B_{1} sin λx, T = A_{2} cos λct +B_{2} sin λct .[/latex]

Since y = X · T , we have

[latex]y(x, t) = (A_{1} cos λx + B_{1} sin λx)(A_{2} cos λct +B_{2} sin λct).[/latex] (9.9)

From the condition y(0, t) = 0, it follows that [latex]A_{1}(A_{2} cos λct + B_{2} sin λct) = 0[/latex]. This condition is satisfied by [latex]A_{1} = 0[/latex]. Then

[latex]y(x, t) = B_{1} sin λx(A_{2} cos λct + B_{2} sin λct).[/latex]

We now set

[latex]B_{1}A_{2} = a, B_{1}B_{2} = b,[/latex]

and it follows that

[latex]y(x, t) = sin λx(a cos λct +b sin λct).[/latex] (9.10)

From the condition y(l, t) = 0, it follows that sin λl = 0. This happens if

[latex] λ l = nπ or λ = \frac{nπ}{l}.[/latex] (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:

[latex]y_{n}(x, t) = sin \frac{nπx}{l} \left(a_{n} cos \frac{nπct}{l} +b_{n} sin \frac{nπct}{l}\right). [/latex] (9.12)

Because y(x, 0) = 0, all [latex]a_{n}[/latex] = 0, we have

[latex]y_{n}(x, t) = b_{n} sin \frac{nπx}{l} sin \frac{nπct}{l}.[/latex] (9.13)

By differentiation of (9.13), we get

[latex]\frac{∂y_{n}}{∂t} = b_{n} \frac{nπc}{l} sin \frac{nπx}{l} cos \frac{nπct}{l}.[/latex] (9.14)

For linear differential equations, the superposition principle holds, so that the entire solution looks as follows:

[latex]\frac{∂y}{∂t}=\sum\limits_{n=1}^{∞}{\frac{nπcb_{n}}{l} sin \frac{nπx}{l} cos \frac{nπct}{l}}.[/latex] (9.15)

Because

[latex]\frac{∂}{∂t}y(x, t) |_{t=0}= g(x),[/latex]

it follows that

[latex]g(x) =\sum\limits_{n=1}^{∞}{\frac{nπcb_{n}}{l} sin \frac{nπx}{l}}.[/latex] (9.16)

The Fourier coefficients then follow by

[latex]\frac{nπcb_{n}}{l}= \frac{2}{l} \int\limits_{0}^{l}{g(x) sin \frac{nπx}{l}dx}[/latex]

or

[latex]b_{n} = \frac{2}{nπc} \int\limits_{0}^{l}{g(x) sin \frac{nπx}{l}dx}[/latex]

By inserting (9.18) into (9.13), we obtain the final solution for y(x, t):

[latex]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}}.[/latex] (9.19)

Question 9.3: Find the transverse displacement of a vibrating string of le...

Related Answered Questions

Which trajectory of the mass of a mathematical pendulum yields a pendulum period that is independent of the amplitude? ...

Verified Answer:

Given the function (a) Sketch the function. (b) Determine its Fourier series. ...

Verified Answer:

Find the Fourier series of the function f (x) = 4x, 0 ≤ x ≤ 10, with period 2l = 10, l = 5. ...

Verified Answer: