Question

A thin string of length L = 3m is pinned at the two ends. Based on the material linear density ρ and on the traction force T, it can be assumed that √ T/ρ = 2 m/s. The string is initially straight when, at time t = 0, it is tapped in two narrow intervals (1 − 1/8, 1 + 1/8) and (2 − 1/8, 2 + 1/8) with opposite velocities +1m/s and −1m/s, respectively. i) Derive the field equation of the vibrating string for the string displacement u(x, t) and define the complete set of boundary and initial conditions required to solve the problem. ii) Assume that u(x, t) can be find in the form u(x, t) = X(x)T(t) and show how the initial-boundary value problem defined in i) can be reformulated in terms of the functions X(x) and T(t). iii) Discuss how the boundary conditions impose precise restrictions on the solution of the problem. iv) Solve the problem for the specific case at hand and find the expression of u(x, t) which satisfies the full set of field, initial and boundary conditions.

Answer 1

i)

The field equation is the 1D wave equation:

Here

The boundary conditions:

The first condition is the boundary condition. It basically means that the displacement at the ends of the string is 0. This holds because, the string is pinned at two ends.

The second condition is an initial condition. Since the string was tapped (like a piano string) and not pulled (guitar string), the displacement of the string initially is 0 everywhere.

The third one defines a functional form of the initial velocity. The string was tapped betweeb 1-1/8 and 1+1/8 and between 2-1/8 and 2+1/8

The velocity exists only at these locations. The function is defined as:

ii)

The separation of variables:

Re-formulating the differential equation:

Dividing both sides by XT,

Introducing the separation constant -k2

So the two differential equations:

................(1)

And

..................(2)

iii)

The solutions of eq (1):

and the solutions of eq (2):

From the X xolutions we can ignore the cosine term. The displacement at the ends is 0. For the cosine term, at x=0, it will always go to 1. So we can ignore this. So we are only left with the sine solution.

Also the boundary condition is:

So,

therefore,

So, where n is an integer

since L is 3m,

These are the only allowed values of k

iv)

Now applying the condition:

Again, we will ignore the cosine solution since it goes to 1 at t=0

So,

So the basis solution:

Therefore, the general solution:

The initial condition:

Let

So,

This looks like a Fourier sine series.

So,

These are the constants that define the final solution.