# Deriving 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?

**Cite as:**Deriving Exponential Decay from Damping Forces.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/deriving-exponential-decay-from-damping-forces/