# Damped Harmonic Oscillators

**Damped harmonic oscillators** are vibrating systems for which the amplitude of vibration decreases over time. Since nearly all physical systems involve considerations such as air resistance, friction, and intermolecular forces where energy in the system is lost to heat or sound, accounting for damping is important in realistic oscillatory systems. Examples of damped harmonic oscillators include any real oscillatory system like a yo-yo, clock pendulum, or guitar string: after starting the yo-yo, clock, or guitar string vibrating, the vibration slows down and stops over time, corresponding to the decay of sound volume or amplitude in general.

Mathematically, damped systems are typically modeled by simple harmonic oscillators with **viscous damping** forces, which are proportional to the velocity of the system and permit easy solution of Newton's second law in closed form. These are second-order ordinary differential equations which include a term proportional to the first derivative of the amplitude. As described below, the magnitude of the proportionality describes how quickly the vibrations of the damped oscillator damp down to nothing.

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## Exponential Decay from Damping Forces

Damping forces are often due to motion of an oscillatory system through a fluid like air or water, where interactions between the molecules of the fluid (e.g. air resistance) become important. At low velocities in non-turbulent fluid, the damping of a harmonic oscillator is well-modeled by a viscous damping force \(F_d = -b \dot{x}\). Adding this term to the simple harmonic oscillator equation given by Hooke's law gives the equation of motion for a viscously damped simple harmonic oscillator.

\[m\ddot{x} + b \dot{x} + kx = 0,\]

where \(b\) is a constant sometimes called the **damping constant**.

Solutions should be oscillations within some form of damping envelope. Ansatz an exponential damping envelope:

\[x(t) = Ae^{-gt} e^{iat} = Ae^{t (ai-g)} = Ae^{rt}.\]

where \(A\) is some constant and \(r = ai-g\) will be found. Plugging this ansatz into the equation of motion yields:

\[(mr^2 + b r + k)Ae^{rt} = 0 \implies r^2 + \frac{b}{m} r + \frac{k}{m} = 0,\]

which is a quadratic equation in \(r\) with solutions:

\[r = -\frac{b}{2m} \pm i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }.\]

Note that there are two roots \(r\) to the quadratic as long as the imaginary part is nonzero, corresponding to the two general solutions:

\[x(t) = Ae^{-\frac{b}{2m} t} e^{ i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t} + Be^{-\frac{b}{2m} t} e^{ -i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t}.\]

These solutions in general describe oscillation at frequency \(\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }\) within a decay envelope of time-dependent amplitude \(e^{-\frac{b}{2m} t}\).

Depending on the values of \(m\), \(k\), and \(\gamma\), the solution exhibits different types of behavior:

\(b^2 < 4km\): **Underdamping**

Underdamped solutions oscillate rapidly with the frequency and decay envelope described above. For objects with very small damping constant (such as a well-made tuning fork), the frequency of oscillation is very close to the undamped natural frequency \(\omega_0 = \sqrt{\frac{k}{m}}\).

\(b^2 = 4km\): **Critical damping**

This case corresponds to the vanishing of the frequency \(\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }\) described previously. At this level of damping, the solution \(x(t)\) most rapidly approaches the steady-state amplitude of zero. Larger amounts of damping (see overdamping) cause the solution to more slowly approach zero as it moves slowly through the damping fluid, whereas smaller amounts of damping cause the solution to oscillate more rapidly around zero. Notably, solutions at critical damping do not oscillate.

If the frequency\(\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }\) vanishes, the two linearly independent solutions can be written:

\[x(t) = Ae^{-\frac{b}{2m} t} + Bte^{-\frac{b}{2m} t} .\]

\(b^2 > 4km\): **Overdamping**

In the overdamped case, the frequency \(\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }\) becomes imaginary. As a result, the oscillatory terms \(e^{i\omega t}\) and \(e^{-i\omega t}\) become growing and decaying exponentials \(e^{|\omega| t}\) and \(e^{-|\omega |t}\). Overdamped solutions do not oscillate and instead slowly decay towards equilibrium.

A \(2 \text{kg}\) mass attached to a spring of spring constant \(k = 10 \text{ N}/\text{m}\) oscillates through a fluid that exerts a damping force \(F_d = -(4 \text{ N}\cdot\text{s}/\text{m})\: v\) on the mass, where \(v\) is the velocity of the mass. Which of the following correctly describes the oscillatory behavior of the system?

## Amplitude Attenuation from Exponential Decay

Consider the equation of motion of the underdamped harmonic oscillator:

\[x(t) = Ae^{-\frac{b}{2m} t} e^{ i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t} + Be^{-\frac{b}{2m} t} e^{ -i\sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }t}.\]

This solution describes rapid oscillation within an envelope of exponentially decaying envelope. The amplitudes of the critically damped and overdamped harmonic oscillators similarly decay exponentially.

Several parameters are used throughout physics and engineering literature to describe how the amplitude of a damped harmonic oscillator decays over time. The most mathematically straightforward parameter is the **\(1/e\) decay time**, often denoted as \(\tau\). Suppose a damped harmonic oscillator starts at amplitude \(x_0\); then the amplitude of the damping envelope is \(x_0 e^{-\frac{b}{2m} t}\). The \(1/e\) decay time is defined as the time \(\tau\) for which the amplitude has decreased to \(x_0 / e \approx .368 x_0\). This is equivalent to the exponent in the decay envelope taking value \(-1\), i.e.:

\[-\frac{b}{2m} \tau = -1 \implies \tau = \frac{2m}{b} .\]

A \(10 \text{ kg}\) mass is attached to a spring of spring constant \(10 \text{ N}/\text{m}\). The entire system is submerged in water, which exerts a viscous damping force on the mass \(F_d = -(2 \text{ N}\cdot \text{s}/\text{m}) \:v\). The mass is pulled so that the spring is displaced from equilibrium by \(.1 \text{ m}\) and is released. Find the \(1/e\) decay time of oscillation in seconds.

A more sophisticated parameter is the **quality factor** \(Q\):

\[Q = \frac{\text{energy stored}}{\text{energy dissipated per radian}}.\]

As a mnemonic for understanding and remembering the name, a high *quality* crystal will ring for a very long time when struck. Damped harmonic oscillators with large quality factors are underdamped and have a slowly decaying amplitude and vice versa. Critical damping occurs at \(Q = \frac12\), marking the boundary of the two damping regimes.

## What is the quality factor of a damped harmonic oscillator in terms of \(k\),\(m\), and \(b\)?

Solution:

The stored energy in the damped harmonic oscillator is the "spring potential energy": \[E(t) = \frac12 kA(t)^2\] where \(A(t)\) is the amplitude of the harmonic oscillator. Recalling that the damped harmonic oscillator has a \(e^{-\frac{b}{2m} t}\) decay envelope, this is equal to: \[E(t) = \frac12 k A^2 e^{-\frac{b}{m} t} = E_0 e^{-\frac{b}{m} t}.\]

The energy dissipated per radian is: \[\Delta E = \left|\frac{dE}{dt}\right| \Delta t,\] with \(\Delta t\) giving the time it takes to oscillate through one radian, equal to \(\frac{1}{\omega}\).

The derivative is given by \(\frac{dE}{dt} = -\frac{b}{m} E(t)\), so the energy dissipated is: \[\Delta E = \frac{b}{m \omega} E,\]

and finally the quality factor is: \[Q = \frac{E}{b/(m\omega) E} = \frac{m\omega}{b},\] where \(\omega = \sqrt{\frac{k}{m} - \frac{b^2}{4m^2} }\) is the frequency of the damped harmonic oscillator. For highly underdamped systems, \(\omega \approx \sqrt{\frac{k}{m}}\) and the quality factor is \(Q \approx \frac{\sqrt{km}}{b}\). From this, it is apparent that critical damping occurs at \(Q = \frac12\) by squaring both sides and comparing to the criterion for critical damping.

A last metric that is more common in engineering literature for describing amplitude attenuation in damped oscillators is the **logarithmic decrement** \(\delta\), defined as the natural logarithm of the ratio of amplitudes of two successive peaks for an underdamped harmonic oscillator.

Challenge problem: show that \(\delta = \frac{\pi}{Q}\).

It is important to note that the viscous damping model is a good model only for intermolecular forces in certain fluids. It is *not* a good model for **dry friction**, the usual friction force from rubbing against solid objects governed by the equation \(F_f = \mu N\), with \(\mu\) the coefficient of friction and \(N\) the normal force. Interestingly, simple models of dry friction are solvable and demonstrate a *linear* damping envelope rather than an exponential.

## Problems and Phenomena

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.

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.

## 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 \(s\):

## References

[1] D. Kleppner and R. Kolenkow, *An Introduction to Mechanics*. McGraw-Hill, 1973.

[2] By V4711This vector graphics image was created with Adobe Illustrator.This file was derived from RLC series circuit.png: - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=29120181.

[3] DeCross, M. *Characterization of Cryogenic Deuterium-Tritium Target Motion*. Laboratory for Laser Energetics, 2011, http://www.lle.rochester.edu/media/publications/high_school_reports/documents/hs_reports/2011/DeCross.pdf

**Cite as:**Damped Harmonic Oscillators.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/damped-harmonic-oscillators/