Week 3

More degrees of freedom (uncoupled)

Examples of SHO’s (torsion balance, rigid rod, complicated pendula)

Damped SHO

You can use the following mathematica file to play with Lissajous figures and also the damped simple harmonic oscillator: https://jhubisz.expressions.syr.edu/phy360/wp-content/uploads/sites/5/2016/09/Week-3-1.nb

Multiple Degrees of Freedom

The first thing we discuss this week is a revisit of the idea of having systems with more than one degree of freedom. Recall the video we watched last week (you can download it in the “course documents” section on the right hand side of the page), where there were multiple bowling ball pendula with varying lengths that were chosen to create great visual effects when you view their superposition (combined oscillation given particular initial conditions – their effects don’t actually add together visually). In the beginning of the first lecture this week, we look at the simplified case where there are two masses on springs, and then we view the electrical analog using two function generators whose frequencies we can manipulate and also view utilizing an oscilloscope.

You can use this weeks mathematica file to play with dual oscillators and make your own Lissajous figures.

We will soon see (in a couple of weeks) that coupled oscillators are extraordinarily interesting in terms of the phenomena that can occur. As an example, I showed in class the Wilburten pendulum, where a mass can both spin and bounce on a spring, and where the energy slowly moves between the rotational degree of freedom and the translational (vertical) degree of freedom.

Also, you might enjoy the following video, which emphasizes the role in coupling between metronome degrees of freedom, where the metronomes become “phase locked” due to the coupling: https://www.youtube.com/watch?v=5v5eBf2KwF8

Damping:

Secondly, we began to discuss the phenomenon of damping. Damping is present in nearly all systems (certainly all that you are familiar with in day-to-day life). Damping is the result of energy in a system being lost to its environment through various different channels (sound wave energy, friction/viscous stirring, heat, etc…). In many systems, damping occurs through a velocity dependent force term. If the velocity is small, the force is approximately

$\overrightarrow{F}_\text{damp} \approx - b \dot{x}(t)$

*Note – again, the vector arrow on top of the force indicates the importance of the sign – i.e. that the force is in the opposite direction as the velocity. The value of $b$ is determined by physical properties of the system such as air density, temperature, or water density, etc.

In this case, Newton’s 2nd law is:

$\overrightarrow{F} = -k x(t) - b \dot{x}(t) = m \ddot{x}(t)$

and we can write the new equation that includes the damping term as

$\ddot{x}(t)+\gamma \dot{x}(t)+\omega_0^2 x(t) = 0$ , where $\gamma = b/m$ and $\omega_0^2 = k/m$ .

Before we delve into the mathematics of this equation, let us remember the phenomena that occur as a result. We performed a demo in lecture where we placed a mass on a spring, as before, but then placed the mass into a narrow glass container holding water. The water must move around when the mass is moving, but there is a viscous force resisting the motion of a mass that is of the form discussed above. The motion that the mass underwent was oscillation, but with an amplitude that decreased quickly with time. The energy of motion and compression of the spring dissipated into motion of the water (which eventually manifests as an increased temperature of the water).

We can solve this equation generally using complex notation, where we use $z(t)$ in place of $x(t)$ , and we suppose there are solutions like $z(t) = C e^{\rho t}$ . We don’t know what $\rho$ is yet, but we can plug into the equation above, and we see that the differential equation turns into an algebraic equation for $\rho$ :

$\rho^2 + \gamma \rho + \omega_0^2\rho =0$ ,

which is solved for two values of $\rho$ , as expected from the quadratic formula:

$\rho_\pm = - \gamma/2 \pm i \sqrt{ \omega_0^2 - \gamma^2/4 }$

This means that there are two types of solutions to the motion, each with their own coefficient:

$z(t) = C_+ e^{\rho_+ t} + C_- e^{\rho_- t}$ where the two $C$ ‘s are determined by initial conditions, and the $\rho$ ‘s are determined by the physical properties of the system.

There are three possibilities for the physical behavior of the system, and they depend on the ratio $Q = \omega_0/\gamma$ . The value of $Q$ for a system is referred to often as the “Quality factor.” In terms of $Q$ , you can rewrite the $\rho$ ‘s as:

$\rho_\pm = \omega_0 (-1/2/Q \pm i \sqrt{1-1/4/Q^2})$

The three possibilities correspond to whether or not the quantity in the square root in $\rho_\pm$ is positive, zero, or negative.

The three possibilities are

Under-damped (or sub-critical damping). In this case, $Q > 1/2$ and the quantity in the square root is positive. The system has oscillatory solutions, with frequency $\omega' = \omega_0 \sqrt{1-1/4/Q^2}$ . The amplitude of the oscillations decays exponentially. After taking the real part of the complex valued solution, the oscillation is characterized by: $x(t) = A e^{- \omega_0 t/2/Q}\cos (\omega' t + \alpha)$ .
Over-damped (or super-critical damping). In this case, $Q < 1/2$ . In this case, there are no oscillatory solutions, only two damped exponentials. The solution has a fast-damped piece (sometimes referred to as a transient) and a slower damped piece. At small times after excitation, the fast damped piece plays a role, but the slow damped solution is the only remaining one at long times after the system has been displaced from equilibrium. The two exponentials have decay coefficients given by $\lambda_\text{slow} = \gamma/2 - \sqrt{1-4 \omega_0^2/\gamma}$ and $\lambda_\text{fast} = \gamma/2 + \sqrt{1-4 \omega_0^2/\gamma}$ (note that both $\lambda$ ‘s are positive), and the solution can be written as $x(t) = A e^{-\lambda_\text{slow} t}+B e^{-\lambda_\text{fast} t}$ .
Critically damped In this special case, we have $Q = 1/2$ . At first glance, this situation looks strange, in that $\rho_+ = \rho_- = -\gamma/2$ , and you might think that there is only one solution then: $x(t) = A e^{- \gamma/2 t}$ . In fact, there are still two solutions, and we can write $x(t) = (A + B t) e^{-\gamma t/2}$ . You can verify that this solution works by plugging it into the original equation of motion. You can also obtain this solution by a Taylor expansion of the over or underdamped scenarios as I suggested in class, so thought it looks quite different in form, it is not drastically different in terms of the phenomena that occur at this critical point.

Physics 360

Vibrations and Waves

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