Linear differential equations
For a relationship such as \(\ddot{x} = ax\), one can write down:
(note) \(a = -\omega^2\)
If a is negative
\[ x(t) = Ae^{i\omega t} + Be^{-i\omega t} \]
SHM
Hooke’s law
\[ \ddot{x} + \omega^2 x = 0 \]
Note:
If you are given initial position and velocity, \(x_0\) and \(v_0\), then \(x(t) = C\cos{\omega t} + D\sin{\omega t}\) works best as \(C = x_0\) and \(D = v_0/\omega\)
Damped harmonic motion
Given a damping force of the form (including the restitution force from the string):
\[ F = -b\dot{x} - kx \]
gives an implicit F = ma equation of:
\[ \ddot{x} +2\gamma \dot{x} + \omega^2 x = 0 \]
where \(\gamma = b/2m\) and omega is the same as before.
We place the restriction that:
- \(\gamma > 0\)
- \(\omega^2 > 0\)
And defining \(\Omega^2 = \gamma^2 - \omega^2\):
\begin{equation} x(t) = e^{-\gamma t}(Ae^{\Omega t} + Be^{-\Omega t}) \end{equation}
which is the solution to the above homogenous differential equation
Underdamping
If \(\Omega^2 < 0\), then \(\gamma < \omega\) and \(\Omega\) is imaginary. We will define the real \(\tilde{\omega} = i\Omega = \sqrt{\omega^2 - \gamma^2}\).
Equation (1) then gives
\begin{align*}
x(t) & = e^{-\gamma t}(Ae^{i\tilde{\omega}t} + Be^{-i\tilde{\omega}t}) \\
& = e^{-\gamma t}C \cos(\tilde{\omega} t + \phi) \\
\end{align*}
The constants are related by \(A + B = C \cos\phi\), \(A-B=iC\sin\phi\), and in a physical problem, A* = B. The \(e^{-\gamma t}\) factor describes the envelope of the motion but not necessarily the curve that passes throught the extremes.
Overdamping
Opposite of overdamping
\begin{align*}
x(t) & = e^{-\gamma t}(Ae^{i\Omega t} + Be^{-i\Omega t}) \\
& = Ae^{-(\gamma - \Omega)t} + Be^{-(\gamma + \Omega)t}
\end{align*}
If \(\gamma\) is only slightly larger than \(\omega\) (\(\Omega \approx 0^+\)), then we’d have exponential decay according to the ‘e’ coefficient. If \(\gamma » \omega\), then \(\gamma \approx \Omega\) and the first term on the right-most side dominates as it has the less negative exponent. We could approxmiate \(\Omega \approx \gamma(1-\omega^2/2\gamma^2)\). and the behavior goes like \(e^{-\omega^2 t/2\gamma}\)
Critical damping
(Note that we can’t use the solution to the diff eq derived above because they’re identical. The second unique solution is \(te^{-\gamma t}\))
When \(\Omega = 0\) (\(\gamma = \omega\)),
\[ x(t) = e^{-\gamma t}(A+Bt) \]
Motion converges to zero in the quickest way
Driven motion
\[ \ddot{x} + 2 \gamma \dot{x} + ax = C_0e^{i\omega_0 t} \]
We guess that the solution is of the form \(Ae^{i\omega_0 t}\)
Solution:
\[ x(t) =\left(\frac{C_0}{-\omega_0^2 + 2 i \gamma \omega_0 + a}\right)e^{i\omega_0 t} \] + the earlier solution to the homogenous equation
Example:
Take a spring with spring constant, k, a drag force \(F_f = -bv\), and is also subject to a driving force \(F_d(t) = F_d \cos\omega_d t\)
Solution
The force is: \(F(x,\dot{x},t) = -b\dot{x} - kx + F_d\cos{\omega_d t}\). So F = ma gives:
\begin{align*}
\ddot{x} + 2\gamma\dot{x} + \omega^2x & = F\cos{\omega_d t} \\
& = \frac{F}{2} (e^{i\omega_d t} + e^{-i\omega_d t})
\end{align*}
where \(2\gamma = b/m\), \(\omega^2 = k/m\), and \(F =F_d/m\).
The particular to this equation is obtained similar to as above:
\begin{align*}
x(t) & = \left(\frac{F/2}{-\omega^2_d + 2 i \gamma \omega_d + \omega^2}\right)e^{i\omega_d t} \\
& + \left(\frac{F/2}{-\omega^2_d - 2 i \gamma \omega_d + \omega^2}\right)e^{-i\omega_d t} \\
& + e^{- \gamma t}\left(Ae^{\sqrt{\gamma^2 - \omega^2} t} + Be^{-\sqrt{\gamma^2 - \omega^2 } t}\right) \\
\end{align*}
By rationalising the denominators and applying Euler’s identity:
\begin{align*}
x(t) & = \left(\frac{F(\omega^2 - \omega_d^2)}{(\omega^2 - \omega^2_d)^2 + 4\gamma^2\omega_d^2}\right)\cos\omega_d t\\
& + \left(\frac{2F\gamma \omega_d}{(\omega^2 - \omega_d^2)^2 + 4\gamma^2\omega^2_d}\right)\sin\omega_d t\\
& + e^{- \gamma t}\left(Ae^{\sqrt{\gamma^2 - \omega^2} t} + Be^{-\sqrt{\gamma^2 - \omega^2 } t}\right) \\
\end{align*}
We may draw a right angled triangle with \(\omega^2 - \omega_d^2\) as adjacent, \(2\gamma\omega_d\) as opposite and R as their hypotenuse and define \(\phi\) as the corresponding angle to derive:
\[ x_p(t) = \frac{F}{R}cos(\omega_d t - \phi) \]