491422461.
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1.5 The vibrating string
Many physical phenomena might be described by differential equations that are relations
involving rates of change. One example is the vibrations of a perfectly elastic string that can be
represented by the solution of a second order PDE known as the one-dimensional wave equation.
Figure 1.5-1 shows the displacement u(x,t) of a string stretched along the x-axis between x = 0
and x = L. The string is free to vibrate in a fixed plane through the x-axis.
u (x ,t)
0 L x
Figure 1.5-1 String displacement at time t.
The one-dimensional wave equation is given by
2u 2 u
2
= c (1.5-1)
t 2 x 2
The constant c depends on the mass density and the tension of the string. Eq. (1.5-1) can be
interpreted in terms of Newton’s second law applied in the upward direction normal to the x-axis
and in the plane containing the string.
F
a= (1.5-2)
m
Since left hand side of Eq. (1.5-1) is the acceleration of a differential portion of the string, the
right side must be the force per unit mass acting on this string portion. The second derivative of
displacement u with respect to x represents the curvature of the string. The curvature is positive
when the string is concave up hence the force is in the positive or upward direction. If a string is
released from an initial position at rest, those portions of the string where it is concave up will
start to move up and those where it is concave down will start to move down as shown in Figure
1.5-2.
u (x ,t)
0 L x
Figure 1.5-2 Direction of movement for a string released from rest.
Since the one-dimensional wave equation is a PDE with second order in both time and special
coordinates, a total of 4 initial and boundary conditions are required for the solution of the
differential equation. The two boundary conditions are
u(0,t) = 0, for t 0 (1.5-3a)
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and
u(L,t) = 0, for t 0 (1.5-3b)
If an initial displacement f(x) and an initial velocity g(x) are known at a time t = 0, then the two
initial conditions are
u(x,0) = f(x), for 0 x L (1.5-3c)
and
u
(x,0) = g(x), for 0 x L (1.5-3d)
t
The simplest solution that satisfies Eq. (1.5-1) and the boundary and initial conditions is
x ct
u(x,t) = sin cos (1.5-4)
L L
u c x ct 2u c x ct 2
= sin sin 2 = sin cos
t L L L t L L L
u x ct 2u
2
x ct
= cos cos = sin cos
x L L L x 2 L L L
Hence, Eq. (1.5-4) satisfies the one-dimensional wave equation
2u 2 u
2
= c (1.5-1)
t 2 x 2
ct
x = 0, u(0,t) = sin(0)cos =0
L
ct
x = L, u(L,t) = sin()cos =0
L
Hence, Eq. (1.5-4) satisfies the two boundary conditions (1.5-3a) and (1.5-3b). The initial
displacement is then
x x
t = 0, u(x,0) = f(x) = sin cos(0) = sin , for 0 x L
L L
The initial velocity is given by
u c x
t = 0, (x,0) = g(x) = sin sin(0) = 0, for 0 x L
t L L
Thus Eq. (1.5-4) represents the solution where the string starts from rest with initial displacement
x
given by sin for 0 x L. This solution describes the behavior of a string being pushed
L
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from an equilibrium position and released. The only special about this solution is the requirement
that the string must have a particular initial shape which is one arch of a sine function.
Figure 1.5-3 shows the snapshots of the string described by the function u(x,t) = sin(x)cos(t) at
various time. The snapshots are location of the entire string at a fixed value of t. The shape of the
string always remains the same with different amplitude due to the function cos( t). After a time
period t = 2 solved from the relation
ct 2L
= 2 t = = 2 ( since L =1 and c = 1)
L c
the string come back to its initial shape, and its velocity also returns to its initial value. Thus the
2L
string has periodic motion with period . At t = 1 the string is instantaneously at rest since
c
u
= sin(x)sin() = 0. The Matlab program used to generated Figure 1.5-3 is listed in
t
Table 1.5-1.
1 1 1
0.5 0.5 0.5
0 0 0
u
-0.5 -0.5 -0.5
-1 -1 -1
0 0.5 1 0 0.5 1 0 0.5 1
t=0.00 t=0.25 t=0.50
1 1 1
0.5 0.5 0.5
0 0 0
u
-0.5 -0.5 -0.5
-1 -1 -1
0 0.5 1 0 0.5 1 0 0.5 1
t=0.75 t=1.00 t=1.25
1 1 1
0.5 0.5 0.5
0 0 0
u
-0.5 -0.5 -0.5
-1 -1 -1
0 0.5 1 0 0.5 1 0 0.5 1
t=1.50 t=1.75 t=2.00
Figure 1.5-3 String location described by u(x,t) = sin(x)cos(t) at various time.
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__________ Table 1.5-1 Matlab program to plot u = sin(pi*x)cos(pi*t) at various time ___________
% Plot u = sin(pi*x)cos(pi*t) at various t
%
x=0:.02:1;zz=sin(pi*x);
%
% Label the time t for each displacement u, the character vector ax stores the data
%
ax='t=0.00t=0.25t=0.50t=0.75t=1.00t=1.25t=1.50t=1.75t=2.00';
%
% Set y-coordinate from -1 to 1
%
x1=[0 0];y1=[-1 1];
for i=1:9;
t=0.25*(i-1);tt=cos(pi*t);
%
% Extract the time from ax, axi is used to label time
%
ib=1+(i-1)*6;ie=ib+5;
axi=ax(ib:ie);
u=zz*tt;
%
% Divide the plot window into 3 rows and 3 columns using subplot command
%
subplot(3,3,i),plot(x,u,x1,y1)
xlabel(axi);ylabel('u')
end
Normal Modes
A family of curves given by
nx nct
u(x,t) = sin cos , for n = 1, 2, 3, … (1.5-5)
L L
2u 2 u
2
are solutions of = c . These curves are called normal modes of the string. The
t 2 x 2
supposition of the normal modes can describe the motion of a string with a given arbitrary initial
displacement f(x).
The function
nx nct
u(x,t) = b
n 1
n sin
L
cos
L
(1.5-6)
can represent the motion of a string with zero initial velocity with initial displacement f(x):
nx
u(x,0) = b
n 1
n sin
L
= f(x) (1.5-7)
The coefficient bn can then be determined from Fourier series.
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Example 1.5-1
1. Let F and G be arbitrary differentiable functions of one variable. Show that u(x,t) = F(x + ct) +
G(x ct) is a solution to the wave equation
2u 2 u
2
=c
t 2 x 2
2. Using the solution in (1), solve the wave equation with initial data
1 u
u(x,0) = 2 = (x), (x,0) = 0
1 x t
3. Take c = 1, and plot the solution for 15 < x < 15 and t = 0, 1, 2, 4, 6, 8.
Solution
u F G 2u 2 F
2
2 G
2
1. =c c = c + c
t t t t 2 t 2 t 2
u F G 2u 2 F
2
2 G
2
2u 2 u
2
= + c2 = c + c = c
x x x x 2 x 2 x 2 t 2 x 2
2. u(x,t) = F(x + ct) + G(x ct)
Let = x + ct and = x ct so that u(x, t) will become u(, ) where
u(, ) = F() + G()
u dF dG dF dG
= + d =c c d
t d t t d
at t = 0, d = d = dx, therefore
u dF dG dF dG
(x,0) = c c =0 = F(x) = G(x) + K (constant)
t dx dx dx dx
u(x,0) = F(x) + G(x) = G(x) + K + G(x) = (x)
( x) K ( x ct ) K
G(x) = G(x ct) =
2 2 2 2
Similarly
u(x,0) = F(x) + G(x) = F(x) + F(x) K = (x)
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( x) K ( x ct ) K
F(x) = + F(x + ct) = +
2 2 2 2
( x ct ) ( x ct )
u(x,t) = F(x + ct) + G(x ct) = +
2 2
1 1 1 1
For c = 1 and (x) = 2 , u(x,t) = 2
1 x 2 1 ( x t ) 1 ( x t )
2
3. Figure 1.5-4 shows plots of u(x,t) for 15 < x < 15 and t = 0, 1, 2, 4, 6, 8. Table 1.5-2 lists the
Matlab program for Figure 1.5-4.
1 1
0 0
u
u
-1 -1
-20 -10 0 10 20 -20 -10 0 10 20
1 t=0 1 t=1
0 0
u
-1 -1
-20 -10 0 10 20 -20 -10 0 10 20
1 t=2 1 t=4
0 0
u
-1 -1
-20 -10 0 10 20 -20 -10 0 10 20
t=6 t=8
1 1 1
Figure 1.5-4 u(x,t) = 2 for 15 < x < 15 and t = 0, 1, 2, 4, 6, 8.
2 1 ( x t ) 1 ( x t )
2
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__________ Table 1.5-2 Matlab program to plot u(x,t) at various time ___________
% Plot u for example 1.5-1 at various t
%
x=-15:.02:15;
%
% Label the time t for each displacement u, the character vector ax hold the data
%
ax='t=0t=1t=2t=4t=6t=8';tv=[0 1 2 4 6 8];
%
% Set y-coordinate from -1 to 1
%
x1=[0 0];y1=[-1 1];x2=[-15 15];y2=[0 0];
for i=1:6;
t=tv(i);xpt2=(x+t).^2;xmt2=(x-t).^2;
%
% Extract the time from ax, label axi is used for x-axis label for the time
%
ib=1+(i-1)*3;ie=ib+2;
axi=ax(ib:ie);
u=0.5*(1.0./(1+xpt2)+1.0./(1+xmt2));
%
% Divide the plot window into 3 rows and 3 columns using subplot command
%
subplot(3,2,i),plot(x,u,x1,y1,x2,y2)
xlabel(axi);ylabel('u')
end
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