Wilberforce Pendulum

Mike Malenbaum

J. Peter Campbell

Results

Coupling coefficient and resonance

After finding the values for the constants k, m, I and delta, the next step was to determine a relationship between the coupling constant, c, and I.  In order to do that, we had to plot z vs. t and calculate the beat frequency, from which we can calculate c.

Graph 1

There are several things that need to be noted about this graph  First, notice that the degree of energy transfer depends on I.  The degree of energy transfer can be observed by the change in amplitude of the position as a function of time.  For complete energy transfer, there will be a time at which the amplitude is zero.  Remember that according to the theory, this is the point at which wz = wq.  From this graph, it looks like that point occurs near I = 5.05E-5 kg*m^2.

Second, notice that w can be calculated directly from these functions near t = 0, due to the fact that the oscillations are almost purely in the z direction.  We measured w for all I and found it to be w = 5.71 rad/s.

Finally, notice that the beat frequency changes between the two I values.  At I = 5.05E-5 kg*m^2, the beat frequency is significantly shorter (period is longer) than at I = 4.9E-5 kg*m^2.  This is due to the fact that this I is nearer to the resonant condition, at which (theoretically) the beat frequency would go to zero.

Graph 2

Graph 2 demonstrates the transfer of energy between rotational and longitudinal oscillations.  Notice that as the longitudinal oscillations go to zero (~20s), the rotational oscillations are at a maximum.

The next step was to plot c as a function of w, wb, m and I.  Remember, the relationship is:

So, for each I we calculated the wb and plotted c vs I.

Graph 3

As expected, c varies as a function of I.  From this graph, we can also determine the I at which the coupling is the weakest, meaning that the most energy is being transferred between the two modes.  Consequently, that point occurs at I = 5.05E-5 kg*m^2 as we suggested above.

Normal modes

Normal modes are those in which the pendulum will stay indefinitely, not transferring energy between the two degrees of freedom.  From the theory, we know that we can find the normal modes in terms of zo and qo :

So, for any given I, we can set the pendulum in motion in a normal mode.  For example, Graph 4 shows the normal mode z vs t graph for I = 5.156E-5 kg*m^2 and graph 5 shows the |w| vs t for the same system:

Graph 4

Graph 5

This system clearly contains some damping, but the oscillations between the z and q are minimal, which confirms that we found a normal mode.  Further, especially from in Graph 5 one easily calculate the normal frequencies of oscillation.

Normal mode analysis

Graph 6

wz < wq                                                                    z vs. q

Mode 1: w1 = 5.80 rad/s

Graph 7

wz = wq                                                                    z vs. q

Mode 1: w1 = 5.80 rad/s

Mode 2: w2 = 5.63 rad/s

Graph 8

wz > wq                                                                    z vs. q

Mode 2: w2 = 5.63 rad/s

From the theory, we determined that there would be a maximum of two normal modes for the system, since there were two degrees of freedom.  In actuality, the number of normal modes depends on the state of the system.  In this case, only for wz = wq , the point of resonance, are there actually two normal modes.  For the cases where there is only one normal mode it turns out that the eigenfunctions in z and q space are complex, meaning that they have no real manifestation.

Also notice that we have measured the two normal mode frequencies, which were predicted in the theory:

The two modes surround the natural frequency of the system, w, which was calculated above, w = 5.71 rad/s.

Theoretical                                                                                Observed

wb = c/(2*w*) = .152 rad (c=.00631 N/m)                   wb  = .17 rad

Mathematical Modeling

Luckily for us, this system can be analytically solved as described in the theory section.

Using these equations (M=I), and Mathematica, we plotted z(t) vs t and |dq/dt| vs t for the following conditions:

m = .266 kg

I = 5.05 E-5 kg*m^2

The results, one example of which is shown below, closely resemble the experimental data, like Graph 1.

Graph 9

From this graph, we measured w to be roughly 5.64 rad/s.

We even experimented with a little damping, using z*(1-ea*t) vs. t:

Graph 10

We used these solutions to reproduce the c vs. I graph from experiment.  In other words, we measured the beat frequency from the computer-calculated output to compare to our experimental data.  Here are the two plots (theoretical and experimental):

Graph 11 (theory)

Graph 3 (experimental)

The two plots were in very good agreement, which suggests that the experimental values listed above are accurate.

But, despite the fact that we could completely solve the system, Dr. Gfroerer wanted us to pretend like we couldn't, and use Mathematica's NDSolve function.

Using NDSolve, as described in the procedure, we plotted z(t) vs t and |dq/dt| vs t for the following conditions:

k = 8.59    m = .266 kg      I = 5.05 E-5 kg*m^2    c = .00386 (from Graph 3 minima)

delta = .00163    zo = .10 m    qo = 0.

Graph 12

Note that this graph is almost identical to the graph obtained using the analytical solutions.  There is one significant difference.  This solution is based on the measured values of k, delta, whereas the analyical solution is based on the experimentally observed values of w1, w2 and w.  The close approximation again suggests that our values of w1, w2, and w, are accurate.

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