Week 7

Forced oscillations of systems with 2 dof:

Recall the demo! We had the usual two pendula with 1kg masses on strings with a weak spring connecting the two masses. The additional ingredient in the demo on Tuesday was that we attached the piston motor so that it would push sinusoidally on one of the masses. As you showed in your homework, this is equivalent to the addition of a sinusoidal driving force acting on one of the two masses. We then slowly scanned through the various different frequencies on the frequency generator that was controlling the motor. We found that at around 0.67 Hz, the symmetric normal mode was excited with large steady state amplitude. We discovered a resonance of the two-body system, and it coincides with a normal mode excitation! We then explored higher frequencies, and we found at at around 0.77 Hz, the second asymmetric normal mode was excited!

We can see this behavior as arising from the equations of motion, as always. The equations of motion are now:

$\vec{F}_A = - m \omega_0^2 x_A - k (x_A - x_B) + F_0 \cos \omega t = m \ddot{x}_A$ $\vec{F}_B = - m \omega_0^2 x_B - k (x_B-x_A) = m \ddot{x}_B$

Note that the motor was only attached to the first pendulum, so there is no driving force acting on mass B.

Now we use the same trick – adding or subtracting the two equations from each other in order to decouple them:

$\ddot{(x_A + x_B)} + \omega_0^2 (x_A +x_B) = F_0/m \cos \omega t$ $\ddot{(x_A - x_B)} + (\omega_0^2 + 2 k/m) (x_A-x_B) = F_0/m \cos \omega t$

So the mathematics leads us to two simple harmonic oscillators, both driven by the same driving force. When $\omega$ coincides with $\omega_0$, you expect to find resonance for the first (symmetric) mode of oscillation. Similarly, when $\omega$ is close to the frequency $\omega_- = \sqrt{ \omega_0^2 + 2 k/m }$ , you expect to be exciting the resonance for the anti-symmetric normal mode. This is exactly what we saw in the demonstration! We also used the frequencies of the normal modes to determine $k$ .

Normal Modes with 2 dof:

We continued our discussion about normal modes of oscillation on Thursday, wrapping up the discussion using a Mathematica demonstration of computing the normal modes of oscillation in a more general case without obvious symmetry. In this case, there is a procedure. The mathematica file that we used is linked here:

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/10/Week-7.nb

The procedure is to first identify the potential energy function. This demonstration was for the “double pendulum” where one pendulum is attached to the bottom of the first, and both have equal length $l$ and equal mass $m$ . The potential energy function we found was given by (in the small angle approximation):

$U(x_1 ,x_2) = \frac{mg}{2l} \left[ 2 x_1^2 + (x_2-x_1)^2 \right]$

From this one function, we can get everything about the behavior of the system when it is perturbed from equilibrium. First we get the forces on each mass by taking the derivative with respect to the two coordinates, then we apply Newton’s 2nd law. The next step is to assume we have identified a normal mode of oscillation of the system, in which case we have

$x_1 = A r \cos \omega t$ $x_2 = A \cos \omega t$

where $\omega$ is the frequency of the normal mode, and $r$ is the ratio of amplitudes when the normal mode is excited. Recall that for the symmetric pendulum, $r = 1$ for the symmetric normal mode, and $r = -1$ for the antisymmetric normal mode.

Plugging this into the equations of motion, we can re-express the equations as algebraic equations that allow us to determine pairs of $\omega$ and $r$ .

Finally we can plot what the normal modes actually look like, as I did in the mathematica file using the Animate command.

We can also then take arbitrary superpositions of the two normal modes, and adjust the amount of amplitude of each of them.

Physics 360

Vibrations and Waves

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