How to set up this problem with driven torsional oscillator?

[zenterix]

Homework Statement

A torsional oscillator comprises a cylinder with moment of inertia $I$ hanging from a light rod with torsional spring constant $k$. The cylinder also experiences a drag torque equal to $-\mu\dot{\theta}$ when moving with angular velocity $\dot{\theta}$. The top of the rod is driven with angular displacement $\phi(t)=\phi_0\cos{\omega t}$.

Find the steady-state solution for $\theta(t)$.

Plot the amplitude $A(\omega)$ and phase $\delta\omega$ of your solution for $\theta(t)$ in (1) as a function of $\omega$.

Relevant Equations

For your plot assume that the natural frequency of oscillation of the system is $\omega_0=1$ and plot three curves on the same plot with $\frac{\mu}{l}=0.25, 1$, and 2. Label your curves to distinguish the three cases.

The "Vibrations and Waves" problem-solving course on MIT OCW has a section on driven harmonic oscillators which can be seen here.

I would like to do the first of the two problems. Unfortunately, there are two issues

1) The latex is not rendering on that website (relatively minor issue, I think I can get parse it).

2) The problem is about a type of oscillator that I am unfamiliar with.

I'd like to show here how I understood the problem to be set up and then how I set up the equations.

My question is if this set up and equations are correct.

I've stated the problem above as I understood it from the website.

Here is how I set it up mathematically.

$$\vec{\tau}=(-k(\theta-\phi)-\mu\dot{\theta})\hat{k}=I\ddot{\theta}\hat{k}\tag{1}$$

$$I\ddot{\theta}+\mu\dot{\theta}+k\theta=k\phi\tag{2}$$

$$\ddot{\theta}+\frac{\mu}{I}\dot{\theta}+\frac{k}{I}\theta=\frac{k}{I}\phi_0\cos{\omega t}\tag{3}$$

In words, the torque experienced by the cylinder has two components. One is due to the angle difference between the cylinder and the rod relative to the equilibrium position, in which this angle difference is zero. This is the $-k(\theta-\phi)$ term.

The other component is drag which is always in the opposite direction to the angular velocity.

If this is correct, this boils down to the same problem as a driven simple pendulum or a driven RLC circuit.

 

[zenterix]

Assuming the problem is set up correctly, a solution is

$$\theta_p(t)=A(\omega)\cos{(\omega t-\alpha)}\tag{4}$$

$$\alpha=\text{arg}(\omega_0^2-\omega^2+i\gamma\omega)\tag{5}$$

$$=\text{arg}\left ( \frac{k}{I}-\omega^2+i\frac{\mu\omega}{I} \right )\tag{6}$$

$$A(\omega)=\frac{f}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}}\tag{7}$$

$$=\frac{\frac{k\phi_0}{I}}{\sqrt{\left ( \frac{k}{I}-\omega^2 \right )^2+\frac{\mu^2}{I^2}\omega^2}}\tag{8}$$

 

[kuruman]

If this is correct, this boils down to the same problem as a driven simple pendulum or a driven RLC circuit.

Precisely. Only the symbols are different.

 

[zenterix]

The problem then asks us to plot amplitude and phase for different values of $\gamma=\frac{\mu}{I}$.

$\gamma$ represents the amount of damping in the system.

We should expect that as we increase $\gamma$ the system at some point becomes overdamped.

Scenario 1: $\frac{\mu}{I}=0.25$

Amplitude

Phase

Scenario 2: $\frac{\mu}{I}=1$

Amplitude

Phase

Scenario 3: $\frac{\mu}{I}=2$

Amplitude

Phase

 

[zenterix]

So how do we interpret these results?

The natural frequency of the system is $\omega_0=\frac{k}{I}=1$. The smaller the damping term $\gamma$ is, the more significant is the phenomenon of resonance which we observe as the peak in amplitude.

Note that this occurs at a driving frequency smaller than the natural frequency

$$\omega_{max}=\sqrt{\omega_0^2-\frac{\gamma^2}{2}}$$

When $\gamma=2$ this max occurs at $\omega=0$.

What about the phase?

The higher the driving angular frequency, the more out of phase the cylinder.

The limiting phase is $\pi$. In the plots from my previous post, this isn't clear because I only plotted to $\omega=5$, but all three have $pi$ as their asymptote.

The speed at which they reach this limiting phase is different.

Here are some plots of all three phases together

Since

$$\alpha=\text{arg}(\omega_0^2-\omega^2+i\gamma\omega)=\tan^{-1}{\frac{\gamma\omega}{\omega_0^2-\omega^2}}$$

Then when $\omega=\omega_0$ the phase is always $\pi/2$.

Here we see what happens to the imaginary part

For $\omega<\omega_0$ we can see that we're in the first quadrant of the complex plane and a higher $\gamma$ means the imaginary part grows faster. $\alpha$ is initially increases relatively faster as a function of $\omega$ and is always above the $\alpha$ for a smaller $\gamma$.

For $\omega>\omega_0$ we are in the second quadrant of the complex plane and as we increase $\omega$ a higher $\gamma$ means the imaginary part is always more negative than the imaginary part for a smaller $\gamma$. Since the real part is the independent of $\gamma$ this means that the argument is always smaller for the larger $\gamma$.

But this is just math. What about the physics?

 

[BvU]

What about the physics?

Perhaps some of the physics pops up when you consider limiting cases
In electronics and in control engineering one often looks at log-log plots

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