Chapter 7
Partial Differential Equations
7.3 Hyperbolic Equations
7.3-1 The Vibrating String Problem.
Figure 7.3-1 The shape of the string and the forces acting on a small portion of it.
Figure 7.3-1 shows the transverse displacement u(x,t) of a string stretched along the x-axis between x = 0 and x = L. The string isfree to vibrate in a fixed plane through the x-axis. The displacement u(x,t) might be obtained from the solution of a PDE that describes the motion of the string as a function of time. The PDE in this case is formulated from the physical law that is the conservation of momentum or force balance. The following assumptions are made to simplify the model
- The string is homogenous with constant mass per unit length .
- The string is perfectly elastic and offers no resistance to bending so that the tension is tangent to the string.
- The transverse vibrations of the strings are small and take place in a plane containing the x-axis, the xu-plane.
Let denotes the magnitude of the string tension that is assumed to be constant throughout the motion since the change in the length of the string is negligible during its motion. Consider a small portion of the string between 2 points A and B located at x and x + x as shown in Figure 7.3-1. Let 1 and 2be the tensions at the points A and B then |1| = |2| = . Making a force balance in the vertical direction with positive pointing upward, we obtain
Fy = may
sin + sin = may= x
For small angle and
sin tan = (x, t) and sin tan = (x + x, t)
(x, t) + (x + x, t) = x
=
As x 0
=
The expression is known as the wave velocity and will be denoted as c, therefore the hyperbolic equation that describes the string vibration is
= c2
An explicit finite difference equation using a second order central differencing for both the space and time derivatives is known as the midpoint Leapfrog method
= c2
where the subscript i indicates x value and the superscript n indicate t value.
Figure 7.3-2 A computational diagram for the midpoint Leapfrog method.
The expression cis known as the Courant number and will be denoted as Co, hence
= Co2( + ) + 2(1 Co2)
Since the best solution is obtained for the value of Co equal to one, we choose x and t so that c= 1. Hence
= +
Example 7.3-1 ______
A banjo string is 80-cm long with a mass of 1.0 gm. It is stretched with a tension of 3.8580107 dyne (gmcm/s2). At a point 20 cm from one end it is pulled 0.6 cm from the equilibrium position and then released. Find the displacements along the string as a function of time. Use x = 10 cm. How long does it take to complete one cycle of motion? From this, compute the frequency of the vibration.
Solution
The string vibration is described by the wave equation
= c2
where c = = = 5.5556104 cm/s
In finite difference form, the wave equation becomes
= Co2( + ) + 2(1 Co2)
Since the best solution is obtained for the value of Co equal to one, we choose t so that c= 1. Hence
= +
and t = x = 10/5.5556104 = 1.810-4 sec
We need the values of u from the previous two times step to evaluate. Therefore the first time step requires the initial velocity, , of the string so that we can apply the central difference approximation
=
For this problem, the string is released from rest. Hence
= 0 =
For t = 0 (n = 0)
= + = + = 0.5( + )
For time level n = 1 and higher
= +
Table 7.4-1 lists the MATLAB program and the computational results that show the cyclic nature of the wave equation as shown in Figure 7.4-1. The original displacements are reproduces every 16 time steps. The frequency of the vibration is then
f = = = 347.2 cycles/sec
Table 7.4-1 An example of the explicit method for hyperbolic equation ------
%
% Example 7.4-1, wave equation with Midpoint Leapfrog method
%
un=[0 0.3 0.6 0.5 0.4 0.3 0.2 0.1 0];
unp1=un;ix=length(un);ix1=ix-1;
for i=2:ix1
unp1(i)=.5*(un(i-1)+un(i+1));
end
for n=0:20
fprintf('%3.0f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f %5.2f\n',n,un)
unm1=un;un=unp1;
for i=2:ix1
unp1(i)=un(i+1)+un(i-1)-unm1(i);
end
end
> ex7d4d1
0 0.00 0.30 0.60 0.50 0.40 0.30 0.20 0.10 0.00
1 0.00 0.30 0.40 0.50 0.40 0.30 0.20 0.10 0.00
2 0.00 0.10 0.20 0.30 0.40 0.30 0.20 0.10 0.00
3 0.00 -0.10 0.00 0.10 0.20 0.30 0.20 0.10 0.00
4 0.00 -0.10 -0.20 -0.10 0.00 0.10 0.20 0.10 0.00
5 0.00 -0.10 -0.20 -0.30 -0.20 -0.10 0.00 0.10 0.00
6 0.00 -0.10 -0.20 -0.30 -0.40 -0.30 -0.20 -0.10 0.00
7 0.00 -0.10 -0.20 -0.30 -0.40 -0.50 -0.40 -0.30 0.00
8 0.00 -0.10 -0.20 -0.30 -0.40 -0.50 -0.60 -0.30 0.00
9 0.00 -0.10 -0.20 -0.30 -0.40 -0.50 -0.40 -0.30 0.00
10 0.00 -0.10 -0.20 -0.30 -0.40 -0.30 -0.20 -0.10 0.00
11 0.00 -0.10 -0.20 -0.30 -0.20 -0.10 0.00 0.10 0.00
12 0.00 -0.10 -0.20 -0.10 0.00 0.10 0.20 0.10 0.00
13 0.00 -0.10 -0.00 0.10 0.20 0.30 0.20 0.10 0.00
14 0.00 0.10 0.20 0.30 0.40 0.30 0.20 0.10 0.00
15 0.00 0.30 0.40 0.50 0.40 0.30 0.20 0.10 0.00
16 0.00 0.30 0.60 0.50 0.40 0.30 0.20 0.10 0.00
17 0.00 0.30 0.40 0.50 0.40 0.30 0.20 0.10 0.00
18 0.00 0.10 0.20 0.30 0.40 0.30 0.20 0.10 0.00
19 0.00 -0.10 -0.00 0.10 0.20 0.30 0.20 0.10 0.00
20 0.00 -0.10 -0.20 -0.10 -0.00 0.10 0.20 0.10 0.00
______
Figure 7.3-3 The string displacements are reproduced every 16 time steps.
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