THE WAVE EQUATION; THE VIBRATING STRING

4. THE WAVE EQUATION; THE VIBRATING STRING

Let a string (for example, a piano or violin string) be stretched tightly and its ends fastened to supports at x = 0 and x = l. When the string is vibrating, its vertical displacement y from its equilibrium position along the x axis depends on x and t. We assume that the displacement y is always very small and that the slope ∂y/∂x of the string at any point at any time is small. In other words, we assume that the string never gets very far away from its stretched equilibrium position; in fact, we do not distinguish between the length of the string and the distance between the supports, although it is clear that the string must stretch a little as it vibrates out of its equilibrium position. Under these assumptions, the displacement y(x, t) satisfies the (one-dimensional) wave equation

2 = 2 2 ∂x . v ∂t

The constant v depends on the tension and the linear density of the string; it is called the wave velocity because it is the velocity with which a disturbance at one point of the string would travel along the string. To separate the variables, we substitute

y = X(x)T (t)

into (4.1) and get (Problem 3.10)

2 = 2 2 = −k X , dx v T dt

634 Partial Differential Equations Chapter 13

or (4.3)

X ′′ +k 2 X = 0, T+k ¨ 2 v 2 T = 0.

We can see from the physical problem why we use a negative separation constant here; the solutions are to describe vibrations which are represented by sines and cosines, not by real exponentials. Of course, if we tried using +k 2 with k real, we would also discover mathematically that we could not satisfy the boundary conditions.

Recall the following notation used in discussing wave phenomena (see Chapter 7, Problem 2.17):

ν = frequency (sec −1 ) ω = 2πν = angular frequency (radians) λ = wavelength

= wave number v = λν

k=

The solutions of the two equations in (4.3) are

sin kvt = sin ωt, (4.4)

cos kvt = cos ωt, and so the solutions (4.2) for y are are

(4.5) y= where ω = kv.

cos kx

cos ωt

Since the string is fastened at x = 0 and x = l, we must have y = 0 for these values of x and all t. This means that we want only the sin kx factors in (4.5), and also we select k so that sin kl = 0 or k = nπ/l. The solutions then become

The particular combination of solu- tions (4.6) that we should take to solve

a given problem depends on the initial conditions. For example, suppose the string is started vibrating by plucking (that is, pulling it aside a small distance

Figure 4.1

h at the center and letting go). Then we are given the shape of the string at t = 0, namely y 0 = f (x) as in Figure

4.1, and also the fact that the velocity ∂y/∂t of points on the string is zero at t = 0. (Do not confuse ∂y/∂t with the wave velocity v; there is no relation between

Section 4 The Wave Equation; the Vibrating String 635

them.) In (4.6) we must then discard the term containing sin (nπvt/l) since its time derivative is not zero when t = 0. Thus the basis functions for this problem are sin (nπx/l) cos (nπvt/l) and we write the solution in the form

The coefficients b n are to be determined so that at t = 0 we have y 0 = f (x), that is,

= f (x).

n=1

As in previous problems, we find the coefficients in the Fourier sine series for the given f (x) and substitute them into (4.7). The result is (Problem 1)

Another way to start the string vibrating is to hit it (a piano string, for example). In this case the initial conditions would be y = 0 at t = 0, with the velocity ∂y/∂t at t = 0 given as a function of x (that is, the velocity of each point of the string is given at t = 0). This time we discard in (4.6) the term containing cos (nπvt/l) because it is not zero at t = 0. Then, for this problem, the basis functions are sin (nπx/l) sin (nπvt/l) and the solution is of the form

Here the coefficients must be determined so that

nπx (4.11)

nπv ∞ nπx

= V (x), ∂t t=0

sin

b n sin

l that is, V (x), the given initial velocity, must be expanded in a Fourier sine series

n=1

(see Problems 5 to 8). Suppose the string is vibrating in such a way that, instead of an infinite series for y, we have just one of the solutions (4.6), say

for some one value of n. The largest value of sin (nπvt/l), for any t, is 1, and the shape of the string then is

Graphs of (4.13) are sketched in Figure 4.2 for n = 1, 2, 3, 4. (The graphs are ex- aggerated! Remember that the displacements are actually very small.) Consider

636 Partial Differential Equations Chapter 13

Figure 4.2

a point x on the string; for this point sin (nπx/l) is some number, say A. Then the displacement of this point at time t is [from (4.12)]

As time passes, this point of the string oscillates up and down with frequency ν n given by ω n = nπv/l = 2πν n or ν n = nv/(2l); the amplitude of the oscillation at this point is A = sin (nπx/l) (see Figure 4.2). Other points of the string oscillate with different amplitudes but the same frequency. This is the frequency of the musical note which the string is producing. (See Chapter 7, Section 10.) If n = 1 (see Figure 4.2), the frequency is v/(2l); in music this tone is called the fundamental or first harmonic. If n = 2, the frequency is just twice that of the fundamental; this tone is called the first overtone or the second harmonic; etc. All the frequencies which this string can produce are multiples of the fundamental. These frequencies are called the characteristic frequencies of the string. (They are proportional to the characteristic values or eigenvalues, k = nπ/l.) The corresponding ways in which the string may vibrate producing a pure tone of just one frequency [that is, with y given by (4.12) for one value of n] are called the normal modes of vibration. The first four normal modes are indicated in Figure 4.2. Any vibration is a combination of these normal modes [for example, (4.9) or (4.10)]. The solution (4.12) (for one n) describing one normal mode, is a characteristic function or eigenfunction.

The waves in Figure 4.2 are called standing waves. The d’Alembert solution of the wave equation (see Problem 1.2) represents traveling waves. Suppose we combine two traveling waves moving in opposite directions as follows:

cos k(x − vt) − cos k(x + vt) = 2 sin kx sin kvt

(by a trigonometry formula). This is one of the solutions (4.5) so we see that this combination of two traveling waves produces a standing wave. Suppose these two traveling waves are moving along a string which is fastened at x = 0 and at x = l. First consider the wave cos k(x + vt) which is moving in the negative x direction toward x = 0. When it reaches x = 0, it will be reflected, and the combination of the incident and reflected waves must equal zero at x = 0 for all t. We see that this is true in (4.15), so the wave cos k(x − vt) is the reflection of − cos k(x + vt). Now consider cos k(x − vt) traveling toward x = l. When it reaches x = l and is reflected, we can verify (Problem 10) that, if k = nπ/l, then the reflection at x = l is − cos nπ l (x + vt). We can think of a wave traveling back and forth between x = 0

Section 4 The Wave Equation; the Vibrating String 637

and x = l, being reflected at each end. The net result as we see from (4.15) is a standing wave.

So far we have been considering problems in which a string is pinned at both ends. We could, instead, have a “free” end; this means free to move up and down along x = 0 or x = l, say by allowing the end to slide along a frictionless track. The mathematical condition for this is ∂y/∂x = 0 at the free end (compare the condition for an insulated face in Section 3). If the x = 0 end is free, we choose the solution containing cos kx (since ∂ ∂x cos kx = −k sin kx = 0 at x = 0). Then, if

the string is pinned at x = l, we want cos kl = 0, so kl = (n + 1 2 )π. Thus the basis

functions when the x = 0 end is free, the x = l end is pinned, and the initial string velocity is zero, are

For a discussion of these functions, see Chapter 7, Section 11 and Problem 11.11.

PROBLEMS, SECTION 4

As in Sections 2 and 3, use a computer to plot your answers. 1. Complete the plucked string problem to get equation (4.9).

2. A string of length l has a zero initial velocity

and a displacement y 0 (x) as shown. (This ini-

tial displacement might be caused by stopping the string at the center and plucking half of it.) Find the displacement as a function of x and t.

Answer: y=

B n sin

cos

, where B n

= (2 sin nπ/4 − sin nπ/2)/n

n=1

3. Solve Problem 2 if the initial displacement is:

4. Solve Problem 2 if the initial displacement is:

5. A string of length l is initially stretched straight; its ends are fixed for all t. At time t = 0, its points are given the velocity V (x) = (∂y/∂t) t=0 as indicated in the diagram (for example, by hitting the string). Determine the shape of the string at time t, that is, find the displacement y as a function of x and t in the form of a series similar to (4.9). Warning: What basis functions do you need here?

« Answer: y= 8hl sin πx sin πvt 1 sin 3πx sin 3πvt + 1 5πx 5πvt 3 − 3 3 sin

sin −··· .

638 Partial Differential Equations Chapter 13

6. Do Problem 5 if the initial velocity V (x) = (∂y/∂t) t=0 is as shown.

3πx 3πvt «

l l +··· . 7. Solve Problem 5 if the initial velocity is:

Answer: y= 2 sin

8. Solve Problem 5 if the initial velocity is

( sin 2πx/l,

0 < x < l/2,

V (x) =

l/2 < x < l. 9. In each of the Problems 1 to 8, find the frequency of the most important harmonic.

10. Verify that, if k = nπ l , then the sum of the two traveling waves in equation (4.15) is zero at x = l, for all t.

11. Verify (4.16) and find a similar formula for a string pinned at x = 0 and free at x = l. Solve Problems 2, 3, and 4, for a string with a free end (a) at x = 0; (b) at x = l.

12. In Sections 2, 3, 4, we have solved a number of physics problems which led to the expansion of a given f (x) in a Fourier sine series. Look at (2.9) and (2.25), temperature in a plate; (3.12), heat flow; (3.26), wave function for a particle in a box; (4.7) and (4.10), displacement of a vibrating string plucked or struck. If we have expanded a given f (x) in a Fourier sine series on (0, l), we can immediately write the corresponding solutions for these six different physics problems on the same

interval. Do this for f (x) = x − x 2 on (0, 1), that is with l = 1. Make computer plots of your results. 13. Do Problem 12 for f (x) = 1 − cos 2x on (0, π).

14. Do Problem 12 for f (x) = x − x 3 on (0, 1).