does the wave equation really work? michael a. karls ball state university november 5, 2005

Does the Wave Equation Really Work?

Michael A. Karls

Ball State University

November 5, 2005

Modeling a Vibrating String

Donald C. Armstead

Michael A. Karls

The Harmonic Oscillator Problem

A 20 g mass is attached to the bottom of a vertical spring hanging from the ceiling.

The spring’s force constant has been measured to be 5 N/m.

If the mass is pulled down 10 cm and released, find a model for the position of the mass at any later time.

Newton’s Second Law

The (vector) sum of all forces acting on a body is equal to the body’s mass times it’s acceleration, i.e. F = ma.*

*The form F=ma was first given by Leonhard Euler in 1752, sixty-five years after Newton published his Principia.

Hooke’s Law

A body on a smooth surface is connected to a helical spring. If the spring is stretched a small distance from its equilibrium position, the spring will exert a force on the body given by F = -kx, where x is the displacement from equilibrium. We call k the force constant.

*This law is a special case of a more general relation, dealing with the deformation of elastic bodies, discovered by Robert Hooke (1678).

The Harmonic Oscillator Model

Using Newton’s Second Law and Hooke’s Law, a model for a mass on a spring with no external forces is given by the following initial value problem:

where proportionality constant 2 = k/m depends on the mass m and spring’s force constant k, u0 is the initial displacement, v0 is the initial velocity, and u(t) is the position of the mass at any time t.

We find that the solution to (1) - (3) is given by

Verifying the Harmonic Oscillator Model Experimentally

Using a Texas Instruments Calculator Based Laboratory (CBL) with a motion sensor, a TI-85 calculator, and a program available from TI’s website (http://education.ti.com/) , position data can be collected, plotted, and compared to solution (3).

The Vibrating String Problem

A physical phenomenon related to the harmonic oscillator is the vibrating string.

Consider a perfectly flexible string with both ends fixed at the same height.

Our goal is to find a model for the vertical displacement at any point of the string at any time after the string is set into motion.

The Vibrating String Model

Let u(x,t) be the vertical displacement of the string at any point of the string, at any time.

Let x = 0 and x = a correspond to the left and right end of the string, respectively.

Assume that the only forces on the string are due to gravity and the string's internal tension.

Assume that the initial position and initial velocity at each point of the string are given by sectionally smooth functions f(x) and g(x), respectively.

x=0 x=a

The Vibrating String Model (cont.)

Applying Newton's Second Law to a small piece of the string, we find that a model for the displacement u(x,t) is the following initial value-boundary value problem:

Equation (5) is known as the one-dimensional wave equation with proportionality constant c2 = T/ related to the string’s linear density and tension T.

Equations (6) - (8) specify boundary and initial conditions.

The Wave Equation

Solving the wave equation was one of the major mathematical problems of the 18th century.

First derived and studied by Jean d’Alembert in 1746, it was also looked at by Euler (1748), Daniel Bernoulli (1753), and Joseph-Louis Lagrange (1759).

The Vibrating String Model (cont.)

Using separation of variables, we find

Checking the Vibrating String Model Experimentally

To test our model, we stretch a piece of string between two fixed poles.

Tape is placed at seven positions along the string so displacement data can be collected at the same x-location’s over time.

The center of the string is displaced, released, and allowed to move freely.

Using a stationary digital video camera, we film the vibrating string.

World-in-Motion software is used to record string displacements at each of the seven marked positions.

Data is collected every 1/30 of a second.

Assigning Values to Coefficients in Our Model

We need to specify the parameters in our model. Length of string: a = 0.965 m. String center: xm= 0.485 m. Initial center displacement:

d = -0.126 m. To find c, we use the fact that

in our solution, the period P in time is related to coefficient c by c = 2a/P. From Figure 1, which shows

the displacement of the center of the string over time, we find that P is approximately 0.165 seconds.

It follows that c should be about 11.70 m/sec.

0.1 0.2 0.3 0.4 0.5 0.6

tsecDisplacement at x 0.485 m

Figure 1

Initial String Displacement and Velocity

For initial displacement we choose the piecewise linear function:

For initial velocity, we take g(x)≡0.

Determining the Number of Terms in Our Model

Using (10) and (11), we can compute the coefficients an and bn of our solution (9).

Graphically comparing the nth partial sum of (9) at t = 0 to the initial position function f(x), we find that fifty terms in (9) appear to be enough.

Model vs. Experimental Results

Figure 2 compares model and actual center displacement over time.

Clearly, the model and actual data appear to have the same period.

However, our model does not attain the same amplitude as the measured data over time.

In fact, the measured amplitude decreases over time.

This physical phenomenon is known as damping.

The next slide compares our model and experiment at each of the seven points on the string over time!

Model: ------

Actual: ------

0.1 0.2 0.3 0.4 0.5 0.6

Figure 2

Model vs. Experimental Results (cont.)

Model: ------

Actual: - - - -

String Model

0.120.225

0.350.485

0.60.725

0.120.225

0.350.485

0.60.725

Figure 3

Model vs. Experimental Results (cont.)

The next 21 slides show our results as “snapshots” in time at 1/30 second intervals.

Dots represent tape positions along the string.

The solid curve represents the model.

Vibrating String

0.2 0.4 0.6 0.8

Vibrating String

0.2 0.4 0.6 0.8

xm0.0333333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.0666667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.1 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.133333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.166667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.2 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.233333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.266667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.3 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.333333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.366667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.4 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.433333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.466667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.5 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.533333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.566667 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.6 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.633333 sec

Vibrating String

0.2 0.4 0.6 0.8

xm0.666667 sec

Vibrating String

That was the last frame!

Vibrating String Model Error

How far are we off? One measure of the error is the mean of the

sum of the squares for error (MSSE) which is the average of the sum of the squares of the differences between the measured and model data values.

We find that over four periods, the MSSE is 0.000890763 m2 or 0.0298457 m.

Revising Our Model

From our first experiment, it is clear that there is some damping occurring.

As is done for the harmonic oscillator, we can assume that the damping force at a point on the string is proportional to the velocity of the string at that point.

This leads to a new model with an extra term in the wave equation.

Revised Model with Damping

Equation (12) is known as the one-dimensional wave equation with damping, with damping factor .

Coefficient c, initial values, and boundary values are the same as before.

Solution to the Revised Model

Again, using separation of variables, we find that the solution to (12)-(15) is

Solution to the Revised Model (cont.)

Note that if g(x)≡0, the RHS of (18) is zero for all n, it follows that

Assigning Values to Coefficients in the Revised Model

For our revised model, we keep the same values for a, c, and initial position and velocity functions f(x) and g(x).

The only parameter we still need to find is the damping coefficient . Using the string center’s period in time of P = 0.165 seconds,

c = 11.70 m/sec, and the fact that 2 = P1, we guess that should be approximately 0.0127 sec/m2.

Once we know , the coefficients in our solution (16) can be found with (17) - (19). Again we use fifty terms in (16).

Unfortunately, our choice of does not produce enough damping in our model.

Through trial and error, we find that = 0.0253 sec/m2 gives reasonable results!

Revised Model vs. Experimental Results

Figure 4 compares model and actual center displacement over time.

With damping included, there appears to be much better agreement between model and experiment!

Model: ------

Actual: ------

0.1 0.2 0.3 0.4 0.5 0.6

Figure 4

Revised Model vs. Experimental Results (cont.)

Model: ------

Actual: - - - -

String With Damping Model

0.120.225

0.350.485

0.60.725

-0.1-0.05

00.050.1

0.120.225

0.350.485

0.60.725

Figure 5

Revised Model vs. Experimental Results (cont.)

Damped Vibrating String

0.2 0.4 0.6 0.8

xm0.0333333 sec

0.2 0.4 0.6 0.8

xm0.0666667 sec

0.2 0.4 0.6 0.8

xm0.1 sec

0.2 0.4 0.6 0.8

xm0.133333 sec

0.2 0.4 0.6 0.8

xm0.166667 sec

0.2 0.4 0.6 0.8

xm0.2 sec

0.2 0.4 0.6 0.8

xm0.233333 sec

0.2 0.4 0.6 0.8

xm0.266667 sec

0.2 0.4 0.6 0.8

xm0.3 sec

0.2 0.4 0.6 0.8

xm0.333333 sec

0.2 0.4 0.6 0.8

xm0.366667 sec

0.2 0.4 0.6 0.8

xm0.4 sec

0.2 0.4 0.6 0.8

xm0.433333 sec

0.2 0.4 0.6 0.8

xm0.466667 sec

0.2 0.4 0.6 0.8

xm0.5 sec

0.2 0.4 0.6 0.8

xm0.533333 sec

0.2 0.4 0.6 0.8

xm0.566667 sec

0.2 0.4 0.6 0.8

xm0.6 sec

0.2 0.4 0.6 0.8

xm0.633333 sec

0.2 0.4 0.6 0.8

xm0.666667 sec

Damped Model Error

We find that over approximately four periods, the MSSE is 0.000296651 m2 or 0.0172236 m.

Modeling a Vibrating Spring

In order to see if there is any other way to reduce the amount of error we are seeing in our models, we repeat our experiment with a long thin spring in place of our string.

Since the spring is “hollow”, we assume damping due to air resistance is negligable.

Therefore, the classic wave equation IVBVP may be a reasonable model.

Assigning Values to Coefficients in the Spring Model

For our spring model, we choose the same initial position and initial velocity functions f(x) and g(x).

For this experiment, a = 1 m, xm = 0.5 m, d = -0.135 m.

The spring’s period in time is about 0.263 seconds, so using the relationship c = 2 a/P, we find c = 7.6 m/sec.

Spring Model vs. Experimental Results

Figure 6 compares model and actual center displacement over time, after shifting our model in time by -0.02 seconds.

There appears to be even better agreement than in the damped case!

Model: ------

Actual: ------

0.2 0.4 0.6 0.8

Figure 6

Spring Model vs. Experimental Results (cont.)

Model: ------

Actual: ------

Spring Model

0.110.24

0.370.50.615

0.7450.87

0.110.24

0.370.50.615

0.7450.87

Figure 7

Spring Model vs. Experimental Results (cont.)

Vibrating Spring

0.2 0.4 0.6 0.8 1

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.0333333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.0666667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.1 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.133333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.166667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.2 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.233333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.266667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.3 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.333333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.366667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.4 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.433333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.466667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.5 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.533333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.566667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.6 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.633333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.666667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.7 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.733333 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.766667 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.8 sec

Vibrating Spring

0.2 0.4 0.6 0.8 1

xm0.833333 sec

Vibrating Spring

Spring Model Error

We find that over approximately three periods, the MSSE is 0.000174036 m2 or 0.0130551 m.

Conclusions and Further Questions

Using “inexpensive”, modern equipment (rope, spring, video camera, and computer software), we’ve been able to show that the wave equation works!

As is often the case in modeling, we had to revise our initial model or experimental setup to get a model that matches reality.

Conclusions and Further Questions (cont.)

How would the model work without “wobbly” poles?

Would a thinner string reduce damping?

What is really going on with the spring?

Would adding an external force to the models reduce error?

Conclusions and Further Questions (cont.)

One Final Question! Did d’Alembert, Euler, Bernoulli, or

Lagrange ever verify these models via experiment?

If so, how?

References

William E. Boyce and Richard C. Diprima, Elementary Differential Equations and Boundary Value Problems (8th ed).

David Halliday and Robert Resnick, Fundamentals of Physics (2cd ed).

David L. Powers, Boundary Value Problems (3rd ed).

Raymond A. Serway, Physics for Scientists and Engineers with Modern Physics (3rd ed).

St. Andrews History of Math Website: http://www-groups.dcs.st-and.ac.uk/~history/

does the wave equation really work? michael a. karls ball state university november 5, 2005

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