**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 w_{z}
= w_{q}.
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,
w_{b},
m and I. Remember, the relationship is:

**
**

So, for each I we calculated the w_{b}
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 z_{o} and
q_{o}
:

** **

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__

w_{z}
< w_{q}
z vs. q

Mode 1:
w_{1} = 5.80 rad/s

__Graph ____7__

w_{z}
= w_{q}
z vs. q

Mode 1:
w_{1} = 5.80 rad/s

Mode 2:
w_{2} = 5.63 rad/s

__Graph ____8__

w_{z}
> w_{q}
z vs. q

Mode 2:
w_{2} = 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 w_{z}
= w_{q}
, 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__

w_{z}
= = 5.68 rad/s
5.71 rad/s

w_{q}**
= = 5.68 rad/s (I=5.05E-5
kg*m^2)
5.71 rad/s**

w** _{b}
= c/(2***w*

** **w** _{1}=
**w

** **w** _{2}=
**w

__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:

w_{1}
= 5.80 rad/s

w_{2}
= 5.63 rad/s

w = 5.71 rad/s

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-e^{a*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 z_{o} = .10
m q_{o}
= 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 w_{1},
w_{2}
and w.
The close approximation again suggests that our values of w_{1},
w_{2},
and w,
are accurate.