Underdamping: $b^2 ~<~ 4mk$

Let's consider the latter case first. We you get oscillations, we call this underdamping. In this case $\lambda$ is a complex number. It's got a real part and an imaginary part.

The real part is $-b/2m$ and we can figure out the imaginary part by writing $\sqrt{b^2 -4mk}$ as

\sqrt{(-1)(4mk-b^2)} ~=~ i \sqrt{(4mk-b^2)}
\end{displaymath} (1.66)

So we can rewrite the solution for $\lambda$ as
\lambda ~=~ {-b\over 2m} \pm i\sqrt{{k\over m}-{b^2\over 4m^2}}
\end{displaymath} (1.67)

The square root is the magnitude of the imaginary part. When $b~=~ 0$, the square root just becomes $\sqrt{k/m}$, the normal frequency of oscillation, so it makes sense to interpret this as a frequency
\omega ~=~ \sqrt{{k\over m}-{b^2\over 4m^2}}
\end{displaymath} (1.68)


\lambda ~=~ {-b\over 2m}\pm i \omega
\end{displaymath} (1.69)

So are solution
x(t) ~=~ A e^{\lambda t} ~=~ A e^{( {-b\over 2m}\pm i \omega )t}
~=~ A e^{ {-b\over 2m}t} e^{\pm i \omega t}
\end{displaymath} (1.70)

The $\pm$ means there are two solutions here. This is just like our earlier use of imaginary numbers to solve the simple harmonic oscillator. Remember eqn. 1.31. The two solutions there were shown to be equivalent to the two solutions in eqn. 1.32. So the same is true here. The above two solutions are equivalent to
x(t) ~=~
~=~ A e^{ {-b\over 2m}t} \{\cos(\omega t),\sin(\omega t)\}
\end{displaymath} (1.71)

This is what we guessed above before we plunged into all this math. You just have an exponential multiplying a sine wave. But now we know the expression precisely. The amplitude $A(t)$ decays as $e^{{-b\over 2m}t}$. What does this constant $-b/2m$ mean? If it's zero, there is no decay at all. If it's big it decays very fast.

Suppose we start with an amplitude of unity, and want to know the time $\tau$ it takes to decay to $1/e \approx .37$ of its original value. We have $A(t) ~=~ e^{{-b\over 2m}t}$ so at $t~=~ 1$ we have

e^{{-b\over 2m}\tau} ~=~ e^{-1}
\end{displaymath} (1.72)

\tau ~=~ {2m\over b}
\end{displaymath} (1.73)

This is often called the the decay time. We can rewrite the solution in terms of this
x(t) ~=~ A e^{ -t/\tau} \{\cos(\omega t),\sin(\omega t)\}
\end{displaymath} (1.74)

As the damping increases, $\tau$ decreases, that is, it damps faster. But also note that as the damping $b$ increases, $\omega$ decreases finally hitting zero. Now we look at what happens past this point.

josh 2010-01-05