Stage: 5 Challenge Level: 
We were sent two very impressive solutions to this fascinating problem. We can only assume that Ben from Kenny and Paul (no location given) are going to go on to do great things mathematically!
Both solutions were formatted in careful detail, so we include the two pdf files directly for Paul and Ben. Ben also included his geogebra file for you to have a look at.
Undamped Equation of Motion (with linear approximation)
No radial acceleration as $F_{rad}$ balanced by tension in string.
Newton's Second Law:
$$\sum_{i}\mathbf{F_i} = \frac{\textrm{d}}{\textrm{d}t}(m \mathbf{v})$$
Considering the tangential direction:
$$F_{tan} = mg \sin x \textrm{ and } a_{tan} = -\ddot{x} \ell$$
$$\Rightarrow -mg \sin x = m\ddot{x}\ell$$
$\sin x = \frac{y}{\ell}$ from trigonometric definitions $\Rightarrow y = \ell \sin x$
For small $x$ ($x < 20^\circ$ from question), $\sin x \approx x$
$$\Rightarrow y = \ell x \Rightarrow \ddot{x} = \frac{\ddot{y}}{\ell}$$
Substituting:
$$mg \frac{y}{\ell} = -m \frac{\ddot{y}}{\ell} \ell$$
$$\frac{mg}{\ell}y = -\ddot{my}$$
$$\ddot{my} + \frac{mg}{l}y = 0$$
$\Rightarrow k = \sqrt{\frac{mg}{\ell}}$ with units $\sqrt{\textrm{kg}}\ \textrm{s}^{-1}$
Inclusion of friction (damped response)
Taking the frictional
forces acting on the pendulum as being proportional and in opposite
direction to the velocity of the bob is a good model in part,
reflecting well the drag forces acting on the bob, which at such
low velocities do exhibit a proportionality to velocity (Stoke's
drag - see Modelling article) and replicating the observed
real-life behaviour of decaying amplitude of oscillations. This
model does not however account well for the frictional forces
acting at the 'hinge point' of the pendulum as these are likely to
be independent of velocity, however as they are also likely to be
fairly insignificant in most cases, the model is still valid.
Deriving the equation of
motion again using Newton's second law and this time including the
drag force (assuming constant of proportionality $\lambda$):
$$-mg \sin x - \lambda
\dot{x} \ell = m\ddot{x}\ell$$
Note the signs are
important here - acceleration is being taken as positive in the
direction of increasing $x$ and so the weight component is negative
as it acts in the opposite direction, as is the drag force as it
always acts to oppose motion.
As before assuming small
$x \Rightarrow \sin x \approx x$ and using $x =
\frac{y}{\ell}$:
$$-mg \frac{y}{\ell} -
\lambda \frac{\dot{y}}{\ell} \ell = m \frac{\ddot{y}}{\ell}
\ell$$
$$ m\ddot{y} + \lambda
\dot{y} + \frac{mg}{\ell} y = 0$$
$$ m\ddot{y} + \lambda
\dot{y} + k^2 y = 0$$
$\therefore \lambda$ has
units $\textrm{kg s}^{-1}$. It will depend on the geometry the
pendulum (bob) and properties of the fluid it is moving through.
Over the range of velocities that would be expected for a pendulum
of reasonable dimensions acting under Earth's gravitational field
it is reasonable to assume $\lambda$ would remain constant.
Assuming the solution to
the equation has form $y = C e^{pt}$ where $C$ is a constant:
$$\Rightarrow m(p^2 C
e^{pt}) + \lambda (p C e^{pt}) + k^2 (C e^{pt}) = 0$$
Taking non-trivial
solutions, i.e. $C \ne 0$ and as in general $e^{pt} \ne 0$:
$$\Rightarrow mp^2 +
\lambda p +k^2 = 0$$
$$ p = \frac{-\lambda \pm
\sqrt{\lambda^2 - 4(m)(k^2)}}{2m}$$
$$ p =
\frac{-\lambda}{2m} \pm \sqrt{\frac{\lambda^2 -
4mk^2}{4m^2}}$$
$$ p =
\frac{-\lambda}{2m} \pm \sqrt{\frac{\lambda^2}{4m^2} -
\frac{k^2}{m}}$$
Assuming $\lambda^2 <
4mk^2$ i.e. $\lambda^2 < \frac{4m^2g}{\ell}$ (this is likely to
be the case in most real life situations and is called
under-damping) the roots to the characteristic equation will be
complex:
$$ p =
\frac{-\lambda}{2m} \pm
i\sqrt{\frac{k^2}{m}-\frac{\lambda^2}{4m^2}}$$
$$ \Lambda =
\sqrt{\frac{k^2}{m}-\frac{\lambda^2}{4m^2}} \Rightarrow p =
\frac{-\lambda}{2m} \pm i \Lambda$$
Therefore the general
solution is:
$$y = C \exp \left[
\left( \frac{-\lambda}{2m} + i \Lambda \right) t \right] + D \exp
\left[ \left( \frac{-\lambda}{2m} - i \Lambda \right) t
\right]$$
$$y = \exp \left[
\frac{-\lambda}{2m} t \right] \left( C \exp \left[ i \Lambda t
\right] + D \exp\left[- i \Lambda t \right] \right)$$
Euler's formula
$e^{i\theta} = \cos \theta + i \sin \theta$:
$$\Rightarrow y = \exp
\left[ \frac{-\lambda}{2m} t \right] \left( C \left[ \cos (\Lambda
t) + i \sin (\Lambda t) \right] + D \left[ \cos (-\Lambda t) + i
\sin (-\Lambda t) \right] \right)$$
$\cos \theta$ is an even
function $\Rightarrow \cos (-\Lambda t) = \cos (\Lambda t)$
$\sin \theta$ is an odd
function $\Rightarrow \sin (-\Lambda t) = -\sin (\Lambda t)$
$$y = \exp \left[
\frac{-\lambda}{2m} t \right] \left( C \left[ \cos (\Lambda t) + i
\sin (\Lambda t) \right] + D \left[ \cos (\Lambda t) - i \sin
(\Lambda t) \right] \right)$$
$$y = \exp \left[
\frac{-\lambda}{2m} t \right] \left[ (C+D) \cos (\Lambda t) +
i(C-D) \sin (\Lambda t) \right]$$
Let $A = C + D$ and $B = i(C-D)$:
$$y = \exp \left[ \frac{-\lambda}{2m} t \right] \left[ A \cos (\Lambda t) +B \sin (\Lambda t) \right]$$
$$\dot{y} = \frac{-\lambda}{2m} \exp \left[ \frac{-\lambda}{2m} t \right] \left[ A \cos (\Lambda t) +B \sin (\Lambda t) \right] + \exp \left[ \frac{-\lambda}{2m} t \right] \left[ - \Lambda A \sin (\Lambda t) + \Lambda B \cos (\Lambda t) \right]$$
$$\dot{y} = \exp \left[ \frac{-\lambda}{2m} t \right] \left[ \left(\Lambda B - \frac{\lambda}{2m}A \right) \cos (\Lambda t) - \left(\Lambda A + \frac{\lambda}{2m}B \right) \sin (\Lambda t) \right]$$
$$
\begin{align}
\ddot{y} = & \frac{-\lambda}{2m} \exp \left[ \left(\Lambda B - \frac{\lambda}{2m}A \right) \cos (\Lambda t) - \left(\Lambda A + \frac{\lambda}{2m}B \right) \sin (\Lambda t) \right] + \\
& \exp \left[ \frac{-\lambda}{2m} t \right] \left[ -\Lambda \left(\Lambda B - \frac{\lambda}{2m}A \right) \sin (\Lambda t) - \Lambda \left(\Lambda A + \frac{\lambda}{2m}B \right) \cos (\Lambda t) \right]\\
\end{align}
$$
$$\ddot{y} = \exp \left[ \frac{-\lambda}{2m} t \right] \left[ \left( \left( \frac{\lambda}{2m} \right)^2 A - 2 \left( \frac{\lambda}{2m} \right) B - \Lambda^2 A \right) \cos (\Lambda t) + \left( \left( \frac{\lambda}{2m} \right)^2 B + 2 \left( \frac{\lambda}{2m} \right) A - \Lambda^2 B \right) \sin (\Lambda t) \right]$$
Substituting back in to $ m\ddot{y} + \lambda \dot{y} + k^2 y = 0$ gives:
$$
\begin{align}
0 =
& m \left\{
\exp \left[ \frac{-\lambda}{2m} t \right] \left[ \left( \left( \frac{\lambda}{2m} \right)^2 A - 2 \left( \frac{\lambda}{2m} \right) B - \Lambda^2 A \right) \cos (\Lambda t) + \left( \left( \frac{\lambda}{2m} \right)^2 B + 2 \left( \frac{\lambda}{2m} \right) A - \Lambda^2 B \right) \sin (\Lambda t) \right]
\right\} + \\
& \lambda \left\{
\exp \left[ \frac{-\lambda}{2m} t \right] \left[ \left(\Lambda B - \frac{\lambda}{2m}A \right) \cos (\Lambda t) - \left(\Lambda A + \frac{\lambda}{2m}B \right) \sin (\Lambda t) \right]
\right\} + \\
& k^2 \left\{
\exp \left[ \frac{-\lambda}{2m} t \right] \left[ A \cos (\Lambda t) +B \sin (\Lambda t) \right]
\right\}
\end{align}
$$
$$
\begin{align}
0 = \exp \left[ \frac{-\lambda}{2m} t \right]
& \cos (\Lambda t)
\left\{
m \left(\frac{\lambda}{2m}\right)^2 A - 2m \left(\frac{\lambda}{2m}\right) \Lambda B - m \Lambda^2 A + \lambda \Lambda B -\lambda \left(\frac{\lambda}{2m}\right) A + k^2 A
\right\} +
\\
& \sin (\Lambda t)
\left\{
m \left(\frac{\lambda}{2m}\right)^2 B + 2m \left(\frac{\lambda}{2m}\right) \Lambda A - m \Lambda^2 B - \lambda \Lambda A -\lambda \left(\frac{\lambda}{2m}\right) B + k^2 B
\right\}
\end{align}
$$
$$
\begin{align}
0 = \exp \left[ \frac{-\lambda}{2m} t \right]
& \cos (\Lambda t)
\left\{
\left(\frac{\lambda^2}{4m}\right) A - \lambda \Lambda B - m \left( \frac{k^2}{m}-\frac{\lambda^2}{4m^2} \right) A + \lambda \Lambda B -\left(\frac{\lambda^2}{2m}\right) A + k^2 A
\right\} +
\\
& \sin (\Lambda t)
\left\{
\left(\frac{\lambda^2}{4m}\right) B + \lambda \Lambda A - m \left( \frac{k^2}{m}-\frac{\lambda^2}{4m^2} \right) B - \lambda \Lambda A - \left(\frac{\lambda^2}{2m}\right) B + k^2 B
\right\}
\end{align}
$$
$$ \begin{align}
0 = \exp \left[ \frac{-\lambda}{2m} t \right]
& \cos (\Lambda t)
\left\{
\left(\frac{\lambda^2}{4m}\right) A - k^2 A + \left( \frac{\lambda^2}{4m} \right) A -\left(\frac{\lambda^2}{2m}\right) A + k^2 A
\right\} +
\\
& \sin (\Lambda t)
\left\{
\left(\frac{\lambda^2}{4m}\right) B - k^2 B + \left(\frac{\lambda^2}{4m} \right)B - \left(\frac{\lambda^2}{2m}\right) B + k^2 B
\right\}
\end{align}
$$
$$ 0= \exp \left[ \frac{-\lambda}{2m} t \right]
\cos (\Lambda t) \left\{ \left(\frac{\lambda^2}{2m}\right) A - \left(\frac{\lambda^2}{2m}\right) A \right\}
+ \sin (\Lambda t) \left\{ \left(\frac{\lambda^2}{2m}\right) B- \left(\frac{\lambda^2}{2m}\right) B \right\}$$
$$ 0 = \exp \left[ \frac{-\lambda}{2m} t \right] \cos (\Lambda t) \left\{ 0 \right\} + \sin (\Lambda t) \left\{ 0 \right\} = 0$$
$\therefore$ solution is valid.
Taking initial conditions $y(0) = Y_0$ and $\dot{y}(0) = 0$:
$$
\begin{align}
y(0) = Y_0 \Rightarrow & Y_0 = \exp(0) \left[ A \cos (0) +B \sin (0) \right]\\
& Y_0 = (1) \left[ A (1) +B (0) \right] \\
& A = Y_0\\
\end{align}
$$
$$
\begin{align}
\dot{y}(0) = 0 \Rightarrow & 0 = \exp(0) \left[ \left(\Lambda B - \frac{\lambda}{2m} Y_0 \right) \cos (0) - \left(\Lambda Y_0 + \frac{\lambda}{2m}B \right) \sin (0) \right]\\
& 0= (1) \left[ \left(\Lambda B - \frac{\lambda}{2m} Y_0 \right) (1) - \left(\Lambda Y_0 + \frac{\lambda}{2m}B \right) (0) \right] \\
& B = \frac{\lambda Y_0}{2m \Lambda}\\
\end{align}
$$
$$\therefore y = \exp \left[ \frac{-\lambda}{2m} t \right] \left[ Y_0 \cos (\Lambda t) + \frac{\lambda Y_0}{2m \Lambda} \sin (\Lambda t) \right]$$
Maths in STEM. Differential equations. chemistry. Individual. Discussion. biology. Simple harmonic motion. engineering. 1st and 2nd order linear differential equations with constant coefficients. physics.
Published July 2010,August 2010.
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