Springs and oscillators

What is the extension of a spring, and what does it mean when e < 0?

Extension $e = L - \ell$, where $L$ is current length and $\ell$ is natural length. When e < 0 the spring is compressed; when e > 0 it is in tension

State Hooke's Law for a spring connecting points A and B (unit vector $\underline{b}$ from A→B).

$\underline{F_{\text{spring}}} = -ke\underline{b} = -k(L - \ell)\underline{b}$. The force pulls B toward A when stretched (e > 0) and pushes B away when compressed (e < 0).

Derive the equation of motion for a mass $m$ on a spring (stiffness $k$) on the x-axis.

Newton's 2nd law: $m\ddot{x} = -kx$, giving $\ddot{x} + \frac{k}{m}x = 0$. This is the SHM (simple harmonic motion) equation with angular frequency $\omega = \sqrt{k/m}$.

How do you find the spring length $L$, extension $e$, and unit vector $\underline{b}$ for a particle at position $\underline{r}$ attached to a spring fixed at point $\underline{A}$?

1. $L = |\underline{r} - \underline{A}|$ — magnitude of the vector from fixed point to particle.

2. $e = L - \ell$.

3. $\underline{b} = \frac{\underline{r} - \underline{A}}{L}$ — unit vector from $\underline{A}$ toward particle. For a particle at $(x, 0)$ fixed at $(-\ell, 0)$: $L = x + \ell$ (using x > -\ell), so $e = x$. For a tilted case, e.g. particle at $(x, y)$ fixed at $(a, b)$: $L = \sqrt{(x-a)^2 + (y-b)^2}$, then $e = L - \ell$ and $\underline{F}_{\text{spring}} = -ke\underline{b}$ as normal.

What is the general solution to $\ddot{x} + \omega^2 x = 0$, and what is the period?

$x(t) = A\cos(\omega t) + B\sin(\omega t)$, or equivalently $R\cos(\omega t + \theta_0)$.

Period $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$.

Amplitude $R = \sqrt{A^2 + B^2}$ is the maximum displacement.

Phase shift $\theta_0$ satisfies $\tan(\theta_0) = -\frac{B}{A}$.

$A$ and $B$ (or equivalently $R$ and $\theta_0$) are determined by initial conditions $x(0)$ and $\dot{x}(0)$.

How does SHM relate to uniform circular motion?

A particle in a circle of radius $R$ at angular speed $\omega$ has $x(t) = R\cos(\omega t + \theta_0)$ and $y(t) = R\sin(\omega t + \theta_0)$ — each coordinate undergoes SHM, offset by $\pi/2$.

Write the equation of motion for a driven spring (no damping) with external force $F_0\cos(\Omega t)$.

$m\ddot{x} + kx = F_0\cos(\Omega t)$. This is an inhomogeneous 2nd-order linear ODE, solved as $x = x_H + x_P$.

Try $x_P = C\cos(\Omega t) + D\sin(\Omega t)$. Substituting gives $C = \frac{F_0}{k - m\Omega^2}$, $D = 0$, so $x_P = \frac{F_0}{k - m\Omega^2}\cos(\Omega t)$.

What is resonance, and what happens to the solution when $\Omega = \omega$?

Resonance is when the driving frequency matches the natural frequency $\omega$. The denominator $k - m\Omega^2 = 0$, so the standard ansatz fails. Instead try $x_P = Ct\cos(\omega t) + Dt\sin(\omega t)$, giving a solution that grows without bound as $t \to \infty$.

Write the damped oscillator ODE and its characteristic equation.

$m\ddot{x} + \gamma\dot{x} + kx = 0$.

Characteristic equation: $m\lambda^2 + \gamma\lambda + k = 0$, giving $\lambda = \frac{-\gamma \pm \sqrt{\gamma^2 - 4mk}}{2m}$.

Describe the four damping cases and their conditions on $\gamma$.

(1) $\gamma = 0$: undamped SHM - $x(t)=A\sin\sqrt{\frac{k}{m}}t+B\cos\sqrt{\frac{k}{m}t}$

(2) 0 < \gamma < \sqrt{4mk}: underdamped — $e^{-at}(C\cos bt + D\sin bt)$.

(3) $\gamma = \sqrt{4mk}$: critically damped — $x = (A + Bt)e^{-t\sqrt{k/m}}$.

(4) \gamma > \sqrt{4mk}: overdamped — two real negative roots, slow exponential decay, no oscillations. - $x(t)=Ae^{-a}+Be^{-b}$ where $a,b$ are two positive constants from char poly.

Why is critical damping $\gamma = \sqrt{4mk}$ desirable in applications?

It is the boundary between oscillatory and non-oscillatory behaviour — the system returns to equilibrium as quickly as possible without overshooting. Important in devices like microphones and car suspensions.

What is a stable equilibrium point $x_0$ of $m\ddot{x} = F(x)$, and what condition determines stability?

$x_0$ is an equilibrium if $F(x_0) = 0$. It is stable if F'(x_0) < 0 (nearby solutions stay close), and unstable if F'(x_0) > 0.

How do you linearise $m\ddot{x} = F(x)$ near a stable equilibrium $x_0$?

Taylor expand:T

$F(x) \approx F(x_0) + F'(x_0)(x - x_0)$. Since $F(x_0) = 0$, set $X = x - x_0$ to get $m\ddot{X} - F'(x_0)X = 0$.

For stable points this gives SHM with $\omega = \sqrt{-F'(x_0)/m}$.

Express the stability condition and oscillation frequency using the potential $V(x)$, where $\underline{F} = -V'(x)$.

Equilibrium:

$V'(x_0) = 0$.

Stable if V''(x_0) > 0 (local minimum of $V$).

Angular frequency $\omega = \sqrt{\frac{V''(x_0)}{m}}$.

Period $= \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{V''(x_0)}}$.