Damped Oscillators - Problem Solving

The equations of the damped harmonic oscillator can model objects literally oscillating while immersed in a fluid as well as more abstract systems in which quantities oscillate while losing energy. The damped harmonic oscillator is a good model for many physical systems because most systems both obey Hooke's law when perturbed about an equilibrium point and also lose energy as they decay back to equilibrium. These two conditions are sufficient to obey the equation of motion of the damped harmonic oscillator.

Show that a circuit with an inductor, capacitor, and resistor in series obeys the damped harmonic oscillator equation.

A series RLC circuit, in which the resistor dissipates energy while the voltages across the capacitor and inductor oscillate [2].

Solution:

According to Kirchoff's laws, the sum of voltages in a closed loop must be zero. Unlike the figure above, we assume that the capacitor has already been charged and there is no external voltage source connected to the circuit. The voltages across each of the resistor (resistance $R$), capacitor (capacitance $C$), and inductor (inductance $L$) depend on the charge $Q$ on the capacitor and current $I$ in the circuit, where $I = -\frac{dQ}{dt}$ if the capacitor is discharging:

\[ \begin{align} V_R &= IR \\ V_C &= -Q/C \\ V_I &= L \frac{dI}{dt} \end{align} \]

noting that the voltage across the capacitor opposes the flow of conventional current. If the sum of voltages is zero, $I = -\frac{dQ}{dt} = \dot{Q}$ requires:

\[ L\ddot{Q} + R \dot{Q} + \frac{1}{C} Q = 0.\]

The time-dependence of the charge on the capacitor thus behaves like a damped harmonic oscillator. This should be expected: LC circuits are oscillatory, and the energy dissipation from the resistor proportional to the current acts as a damping source.

Anatomy of a fusion target, which oscillates on the flexible Zylon stalk [3].

In inertial confinement fusion, powerful ultraviolet lasers are directed at a small capsule containing isotopes of hydrogen, compressing the hydrogen rapidly to induce fusion. Hydrogen fusion targets are mounted on a stalk made of the highly flexible synthetic polymer Zylon. Although this flexibility prevents targets from breaking off the stalk, it also permits the fusion target to undergo damped oscillation when the laser hits it. Oscillation displaces the center of mass of the target, which reduces the efficiency of the lasers and reduces the chance of fusion; therefore, it is highly desirable to achieve both (1) high fundamental frequencies of oscillation, since these are less easily excited and (2) near-critical damping, to reduce oscillation amplitude quickly [3].

(a) Most of the weight of the hydrogen fusion target is in the mass of the plastic (density approximately $\rho = 1000 \text{ kg}/\text{m}^3 $) that contains the hydrogen, which approximately forms a sphere about $10^{-4} \text{ m}$ in diameter. The $1/e$ decay time after targets are excited is about half of a second. Estimate the spring constant and damping constant of the Zylon stalk, assuming that the targets have a fundamental frequency of about $1000 \text{ Hz}$.

(b) Given the spring constant and damping constant found previously, what mass of the target is necessary to achieve critical damping?

Solution:

(a) First, compute the mass of the fusion target from the volume of the sphere and density of the plastic: \[m = \rho V = (1000\text{ kg}/\text{m}^3)(\frac43 \pi (\frac{10^{-4} \text{ m}}{2})^3) = 5.24 \times 10^{-10} \text{ kg} \]

The $1/e$ decay time is $\tau = \frac{2m}{b}$. From this one obtains the damping constant:

\[b = \frac{2m}{\tau} = \frac{2 (5.24 \times 10^{-10} \text{ kg})}{.5 \text{ s}} = 2.10 \times 10^{-9} \text{ kg}/\text{s}.\]

The spring constant can be extracted from the formula for the (angular) frequency:

\[\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2}} = \sqrt{\frac{k}{m} - \frac{1}{\tau^2}}.\]

Rearranging, the spring constant is: \[k = m \left(\omega^2 + \frac{1}{\tau^2} \right) \]

Substituting in all quantities, keeping in mind $\omega = 2\pi (1000 \text{ Hz})$, the spring constant is calculated:

\[k = 2.07 \times 10^{-2} \text{ kg}/\text{s}^2 .\]

(b) Setting the frequency equal to zero with $b$ and $m$ fixed yields the equation for the mass:

\[km - \frac{b^2}{4} = 0 \implies m = \frac{b^2}{4k}\]

Substituting in for $b$ and $k$ yields the mass:

\[m = 5.33 \times 10^{-17} \text{ kg} .\]

The numbers used in this computation are extremely rough, but one physical property that is clear is that it is impossible to achieve both critical damping as well as a high fundamental frequency. Critical damping requires an essentially massless fusion target in comparison to the mass required for a high fundamental frequency. Balancing the effects of underdamping with large excitation from a low fundamental frequency is a difficult engineering challenge in inertial confinement fusion.

Shock absorbers in the suspension system of cars damp vibrations of the chassis. Ideally, to make the ride as smooth as possible, the vibrations of the chassis will be critically damped. Suppose a car hits a speed bump and the chassis is displaced by $1 \text{ cm}$. If the shock absorbers critically damp the resulting vibration, the car weighs $1000 \text{ kg}$, and the damping constant is $b =20000 \text{ kg}/\text{s} $, find the displacement of the chassis over time.

Solution:

The general solution for critical damping is: \[y(t) = (A+Bt) e^{-\frac{b}{2m} t}.\] Using the initial conditions: $y(0) = 0.1 \text{ cm}$, $\dot{y}(0) = 0$ yields:

\[ \begin{align} A &= 0.1 \text{ cm} \\ B &= \frac{bA}{2m} = 1.0 \text{ cm}/\text{s} \end{align} \]

Below, the solution is plotted in units of $\text{cm}$ over time in $s$: $Shock absorbers critically damp the vibration of the car in a fraction of a second.$ Shock absorbers critically damp the vibration of the car in a fraction of a second.