## Restoring Forces

$$\Sigma F = - kx = m \ddot x \label{eq:hookeslawsumsofforces}$$

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

### Detailed Solution of Hooke's Law

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: $$m \frac{dv}{dt} = - k x \label{eq:hookeslaw}$$

#### Another route:

Rewriting Hooke's Law as $$\frac{d^2 x}{dt^2} + \omega^2 x = 0$$ 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...)

### Analogous systems

#### Simple Pendulum

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. $$F = -kx \;\;\Rightarrow \;\; F = -\frac{mg}{L}s$$

#### Other coordinates?

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

$$\ddot{\theta} + \frac{g}{l}\theta = 0$$

## Damping

$$F = m \ddot{x} = -kx - b \dot{x}$$

#### Clean it up:

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

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

$$\alpha = -\beta \pm \sqrt{\beta^2 - \omega_0} = 0$$

There are 3 possibilities now:

1. Overdamped: $\beta \gt \omega_0$
2. Critically Damped: $\beta = \omega_0$
3. Underdamped: $\beta \lt \omega_0$

#### Overdamped

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

#### Critically Damped

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

#### Underdamped

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.

## Damped and Driven

$$F = m \ddot{x} = -kx - b \dot{x} + F_0 \sin \left(\omega t \right)$$

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.

### Resonance

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.

#### Quality Factor

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 $\beta$.

$$Q = \frac{\omega_0}{2 \beta}$$