Week 5

Coupled Simple Harmonic Oscillators

In this week, we begin to address systems with two degrees of freedom. We first looked at a demonstration where there are two 1kg masses suspended from strings of about 50cm each in length. The masses are then connected to each other through a spring with a small spring constant (called $k_c$ , for example). In the demonstration we did the following:

First we displaced only one of the masses from equilibrium, and set the system in motion. As with the Wilberforce pendulum, energy “sloshed” from one of the degrees of freedom to the other in manner reminiscent of “beats” that we saw while superposing waves of nearly the same frequency.
Next, we displaced both masses by an equal amount in the same direction, such that the spring connecting the two masses was not stretched or compressed. After letting it go, the system just oscillated back and forth, with the spring never being compressed.
Finally, we displaced each mass by an equal amount, but in the opposite direction. This time the masses again just moved back and forth in an antisymmetric manner, somewhat similar to #2.

Clearly the phenomena associated with two degrees of freedom is very rich, exhibiting beats, and having two special types of excitations. How do these appear in the mathematics?

First, we must write down Newton’s 2nd law for each of the masses:

$\overrightarrow{F}_A = -m \omega_0^2 x_A - k_c (x_A -x_B) = m \ddot{x}_A$

$\overrightarrow{F}_B = -m \omega_0^2 x_B - k_c (x_B -x_A) = m \ddot{x}_B$

The most important aspect of these equations are that they are coupled equations. That is, the equation for $x_A$ depends on $x_B$ and vice-versa. The two degrees of freedom “talk to each other” through the coupling spring. Indeed, if the coupling spring constant is taken down to zero, the communication is lost, and the two degrees of freedom become uncoupled.

It is not obvious at this point how to approach finding the solutions to this equation, but you can check that if you add (and subtract) the two equations from each other you get equations that look exactly like the original (uncoupled) SHO equations:

$\ddot{x}_+ = - \omega_0^2 x_+$

$\ddot{x}_- = - (\omega_0^2 + 2 k_c/m) x_-$ .

Here we have defined $x_+ = x_A + x_B$ and $x_- = x_A - x_B$ .

We know these equations! So we know their solutions! We have:

$x_+ (t) = A_+ \cos(\omega_+ t + \alpha_+ )$

$x_-(t) = A_- \cos(\omega_- t + \alpha_- )$ .

There are now 4 coefficients that are determined only after inputting a particular initial condition for the positions and velocities of the two masses.

These solutions can now be used to get back the results for $x_A \text{ and } x_B$ :

$x_A = \frac{1}{2} \left[ A_+ \cos (\omega_+ t + \alpha_+ ) + A_- \cos (\omega_- t + \alpha_- ) \right]$ $x_B = \frac{1}{2} \left[ A_+ \cos (\omega_+ t + \alpha_+ ) - A_- \cos (\omega_- t + \alpha_- ) \right]$

To explore this solution, let’s consider a particular initial condition where at t=0, we have $x_A = A$ and $x_B = 0$ and both velocities are zero. This corresponds to $\alpha_+ = \alpha_- = 0$ and $A_+ = A_- = A$ . Then we have

$x_A = \frac{A}{2} \left[ \cos (\omega_+ t) + \cos(\omega_- t) \right]$

$x_B = \frac{A}{2} \left[ \cos (\omega_+ t) - \cos(\omega_- t) \right]$

Superposing these gives the beat formula! Using $\bar{\omega} = \frac{\omega_+ + \omega_-}{2}$ and $\Delta{\omega} = \frac{\omega_- - \omega_+}{2}$ , we have:

$x_A = A \cos(\bar{\omega} t) \cos\left( \frac{\Delta \omega}{2} t \right)$

$x_B = A \sin(\bar{\omega} t) \sin\left( \frac{\Delta \omega}{2} t \right)$

So we see that, in the equations, the motion/energy “slosh” between the two degrees of freedom. Here is the mathematica notebook that you can use to explore the behavior of this very interesting system!

https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-5.nb

Physics 360

Vibrations and Waves

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