\begin{equation} \Sigma F = - kx = m \ddot x \label{eq:hookeslawsumsofforces} \end{equation}

The general solution for a linear restoring force: \begin{equation} x(t) = A \cos \left(\omega t \right) + B \sin \left( \omega t \right) \end{equation} or, if we use the phase constant: $\phi$ \begin{equation} x(t) = C \cos \left(\omega t + \phi \right) \end{equation}

We'd like to actually solve the diff eq, rather than just 'guessing' at a solution. Here is one method:

Start with the Sums of Forces equation: \begin{equation} m \frac{dv}{dt} = - k x \label{eq:hookeslaw} \end{equation}

Rewriting Hooke's Law as \begin{equation} \frac{d^2 x}{dt^2} + \omega^2 x = 0 \end{equation} we could attempt to solve it by assuming that the solution must take the form: $$x(t) = e^{qt}$$, and then do some work to figure out what $q$ is. (we'll save that...)

For small angles, the tangential force on the pendulum bob will be given by: $$F_\textrm{tang} = -mg \sin \theta$$

When $\theta$ is very small ($\theta < .2$ radians), $\sin \theta \approx \theta$. Using the fact that $s = L \theta$, we can also write: $$F_\textrm{tang} = -\frac{mg}{L}{s}$$

This is essentially a restoring force, just like we had for the mass/spring system. $$F = -kx \;\;\Rightarrow \;\; F = -\frac{mg}{L}s$$

A slightly better coordinate to describe the motion of the pendulum might be $\theta$. That would be a slightly more general coordinate system.

\begin{equation} \ddot{\theta} + \frac{g}{l}\theta = 0 \end{equation}Let's add more terms!

$$F = m \ddot{x} = -kx - b \dot{x}$$\begin{equation} \ddot{x} + 2 \beta \dot{x} + w_0^2 x = 0 \label{eq:dampedoscillatoreqofmotion} \end{equation} where $\beta \equiv \frac{b}{2m}$ and $\omega_0 \equiv \sqrt{\frac{k}{m}}$

A general solution would look like: $$x \propto e^{\alpha t}$$

There are 3 possibilities now:

- Overdamped: $\beta \gt \omega_0$
- Critically Damped: $\beta = \omega_0$
- Underdamped: $\beta \lt \omega_0$

If $$\beta \gt \omega_0$$ then the exponent $\alpha$ in our solution $x = e^{\alpha t}$ will be negative and real.

If $$\beta = \omega_0$$ then solution becomes $x(t) = \left(A + A' t \right) e^{-\beta t}$

If $$\beta < \omega_0$$ then we see that $\sqrt{\beta^2 - \omega^2}$ will be *imaginary*.

Plotted here are the 3 different regimes of a damped oscillator.

Let's add more terms, again!

$$F = m \ddot{x} = -kx - b \dot{x} + F_0 \sin \left(\omega t \right)$$So, now we can solve this:
$$m \ddot{x} + kx + b \dot{x} = F_0 \sin \left(\omega t \right)$$
We want our solution to be the sum of the **characteristic** and **particular** solutions:
$$x(t) = x_c(t) + x_p(t)$$
that is, a general solution for the homogeneous part, and a particular solution for the inhomogeneous equation.

It's helpful to imagine the combination of the two terms in a graphical form.

Here we see the change in the $C$ value as a function of drive frequency (in relation to resonant frequency: $\frac{\omega}{\omega_0}$).

Maximize $C(\omega)$ to find the driving frequency $\omega$ that will lead to the largest response.

$$ C = \frac{f_0}{\sqrt{\left(\omega_0^2 - \omega^2 \right)^2 + 4 \beta^2 \omega^2}} $$This graph shows 5 different damped driven oscillators. The differences in each curve is the ratio of the damping constant to the resonant frequency. Notice the change in shape.

The Full Width Half Max measures describes the shape of the curve.

The Quality Factor is the ratio of the resonant frequency, $\omega_0$, and the damping $\beta$.

\begin{equation} Q = \frac{\omega_0}{2 \beta} \end{equation}