Small Oscillations

Formulation of Small Oscillations only consider systems that satisfy the following:

The system has $n -$ independent generalized coordinates, $q$
$V = V (q)$
${\vec{r}}_{i} = {\vec{r}}_{i} (q)$ , alternatively $x_{i} = x_{i} (q)$

Assume that the system undergoes small oscillations around a static equilibrium, $q_{σ} = q_{σ}^{0}$ , $σ = 1, \dots, n$ Then, ${\dot{q}}_{σ} |_{q_{σ}^{0}} = {\ddot{q}}_{σ} |_{q_{σ}^{0}} = 0$ . Hence, $- {(\frac{\partial V}{\partial q_{σ}})}_{q_{σ}^{0}} = 0$ . We can rewrite: $q_{σ} = q_{σ}^{0} + η_{σ}$ .

$L = T - V = \sum_{j} \frac{1}{2} m_{j} {\frac{d x_{j}}{d t}}^{2} - V (q) = \sum_{j} \frac{1}{2} m_{j} (\sum_{σ} \frac{\partial x_{j}}{\partial q_{σ}} {\dot{q}}_{σ}) (\sum_{λ} \frac{\partial x_{j}}{\partial q_{λ}} {\dot{q}}_{λ}) - V (q) = \frac{1}{2} \sum_{j σ λ} m_{j} (\frac{\partial x_{j}}{\partial q_{σ}}) (\frac{\partial x_{j}}{\partial q_{λ}}) {\dot{q}}_{σ} {\dot{q}}_{λ} - V (q)$ . We can think of the inner sum term as the $(σ,λ)-$th element of a symmetric, real, square matrix $M$ . Then, $L = T - V = \frac{1}{2} M_{σ}^{λ} {\dot{η}}_{σ} {\dot{η}}_{λ}$ . Note that $M_{σ λ} = M_{λ σ} = (M_{σ λ})^{*}$ .

We can also write: $V = V (q) = V (q^{0} + η)$ . Taylor expanding, $V = V (q^{0}) + \sum_{σ} \frac{\partial V}{\partial q_{σ}} |_{q_{σ}^{0}} η_{σ} + \frac{1}{2} \sum_{σ} \sum_{λ} (\frac{\partial^{2} V}{\partial q_{σ} q_{λ}} q_{σ} q_{λ}) + O (q^{3}) = V (q^{0}) + \frac{1}{2} \sum_{σ} \sum_{λ} (\frac{\partial^{2} V}{\partial q_{σ} q_{λ}} η_{σ} η_{λ})$ . Let $V_{σ}^{λ} = \frac{\partial^{2} V}{\partial q_{σ} q_{λ}}$ be the potential square, real, symmetric matrix. So, $V = V (q^{0}) + \frac{1}{2} \sum_{σ λ} V_{σ}^{λ} q_{σ} q_{λ}$ . Note that $V_{σ λ} = V_{λ σ} = (V_{σ λ})^{*}$ .

Therefore, $L = \frac{1}{2} \sum_{σ λ} [M_{σ}^{λ} {\dot{η}}_{σ} {\dot{η}}_{λ} - V_{σ}^{λ} η_{σ} η_{λ}] - V (q^{0})$ .

\begin{aligned} 0 & = \frac{d}{d t} \frac{\partial L}{\partial {\dot{η}}_{k}} - \frac{\partial L}{\partial η_{k}} \\ = \frac{d}{d t} (\frac{1}{2} \sum_{σ λ} [M_{σ}^{λ}] ({\ddot{q}}_{σ} {\dot{q}}_{λ} δ_{σ k} + {\dot{q}}_{σ} {\ddot{q}}_{λ} δ_{λ k})) - (\frac{1}{2} \sum_{σ λ} [M_{σ}^{λ}] ({\dot{q}}_{σ} q_{λ} δ_{σ k} + q_{σ} {\dot{q}}_{λ} δ_{λ k})) \end{aligned}

$\frac{\partial L}{\partial {\dot{η}}_{k}} = \sum_{λ} M_{k}^{λ} {\dot{η}}_{λ}$ . $\frac{d}{d t} \frac{\partial L}{\partial {\dot{η}}_{k}} = \sum_{λ} M_{k}^{λ} {\ddot{η}}_{λ}$ . $\frac{\partial L}{\partial η_{k}} = - \sum_{λ} V_{k}^{λ} η_{λ}$ . So, for each $σ$ , the E-L of the Second Kind are,

\begin{aligned} (1) & 0 & = \sum_{λ} (M_{σ}^{λ} {\ddot{η}}_{λ} + V_{σ}^{λ} η_{λ}) . \end{aligned}

Let $z_{σ}$ where $(σ = 1, \dots, n)$ satisfy the same system of equations, that is complex.

\begin{aligned} 0 & = \sum_{λ} (M_{σ}^{λ} {\ddot{z}}_{λ} + V_{σ}^{λ} z_{λ}) . \end{aligned}

Hence, $η_{σ} = ℜ {z_{σ}}$ .

Solve for Normal Mode Solutions

Hence, $z_{σ} = z_{σ}^{0} \exp (i ω_{0} t)$ for $σ \in {1, \dots, n}$ . Then,

\begin{aligned} 0 & = \sum_{λ} (- ω_{0}^{2} m_{σ}^{λ} + V_{σ}^{λ}) z_{λ}^{0} \exp (i ω t) \\ = \sum_{λ} (V_{σ}^{λ} - ω_{0}^{2} m_{σ}^{λ}) z_{λ}^{0} \end{aligned}

Then,

\begin{array}{r} [V_{0}^{λ} - ω_{0}^{2} M_{σ}^{λ}] (\begin{array}{c} | \\ z^{0} \\ | \end{array}) = 0 \end{array}

$det | V - ω_{0}^{2} M | = 0$ . We will have $n$ frequencies, $ω_{0}^{2} = ω_{s}^{2}$ , $s \in {1, \dots, n}$ . Which gives us, $\sum_{λ} (V_{σ}^{λ} - ω_{s}^{2} M_{σ}^{λ}) z_{λ}^{(s)} = 0$ . It can be shown that $ω_{s}^{2} = (ω_{s}^{2})^{*}$ and $z_λ^(s)/z_p^(s) = $real. So, $z_{λ}^{(s)} = ρ_{λ} (s) \exp (i ϕ_{s})$ , for $ρ_{λ} (s)$ real and $ϕ_{s}$ is a common phase.

$\sum_{σ} z_{σ}^{(s) *} \sum_{λ} (V_{σ}^{λ} - ω_{S}^{2} M_{σ}^{λ}) z_{λ}^{(s)} = 0$ .

\begin{aligned} ω_{S}^{2} \sum_{σ, λ} z_{σ}^{(s)} M_{σ}^{λ} z_{λ}^{(s)} & = \sum z_{σ}^{(s) *} V_{σ}^{λ} z_{λ}^{(s)} \\ ω_{S}^{2} & = \frac{\sum z_{σ}^{(s) *} V_{σ}^{λ} z_{λ}^{(s)}}{\sum_{σ, λ} z_{σ}^{(s) *} M_{σ}^{λ} z_{λ}^{(s)}} \\ (ω_{S}^{2})^{*} & = \frac{\sum z_{σ}^{(s)} V_{σ}^{λ} z_{λ}^{(s) *}}{\sum_{σ, λ} z_{σ}^{(s)} M_{σ}^{λ} z_{λ}^{(s) *}} \\ = \frac{\sum z_{σ}^{(s) *} V_{σ}^{λ} z_{λ}^{(s)}}{\sum_{σ, λ} z_{σ}^{(s) *} M_{σ}^{λ} z_{λ}^{(s)}} \\ = ω_{S}^{2} . \end{aligned}

Recall from last time

$\sum_{λ} (M_{σ}^{λ} {\ddot{η}}_{λ} + V_{σ}^{λ} η_{λ}) = 0, \forall σ \in {1, \dots, n}$ $η_{λ} = ℜ {z_{λ}}$
$\sum_{λ} (M_{σ}^{λ} {\ddot{z}}_{λ} + V_{σ}^{λ} z_{λ}) = 0, \forall σ \in {1, \dots, n}$
(Normal mode solutions) $z_{λ} = z_{λ}^{0} \exp (i ω_{0} t), λ = 1, \dots$ . $\sum_{λ} (V_{σ}^{λ} - ω_{0}^{2} M_{σ}^{λ}) z_{λ}^{0} = 0, \forall σ \in {1, \dots, n}$ .
$| V_{σ}^{λ} - ω_{0} M_{σ}^{λ} | = 0$ . The characteristic frequencies are real and $z_{λ}^{(s)}, z_{π}^{(s)}, \dots$ all have the same phase
For each frequency, $\sum_{λ} (V_{σ}^{λ} - ω_{S}^{2} M_{σ}^{λ}) z_{λ}^{(S)} = 0, \forall σ \in {1, \dots, n}$ .
Plugging in this shared complex phase: $\sum_{λ} (V_{σ}^{λ} - ω_{S}^{2} M_{σ}^{λ}) ρ_{λ}^{(s)} \exp (i ϕ_{S}), \forall σ \in {1, \dots, n}$ . $\sum_{λ} (V_{σ}^{λ} - ω_{S}^{2} M_{σ}^{λ}) ρ_{λ}^{(s)}, \forall σ \in {1, \dots, n}$ . Boundary conditions/ initial conditions can be used to find the phase. Writing this in vector form, $ρ_{λ}^{(s)} \Rightarrow ρ^{(s)}$ . One can show that $\sum_{λ, σ} ρ_{σ}^{(t)} M_{σ}^{λ} ρ_{λ}^{(s)} = ρ^{(t) T} M ρ^{(s)} = δ_{s, t} ρ^{(t) T} M ρ^{(s)}$ Thus, we can make them orthonormal to the mass matrix. At that point: $ρ^{(t) T} M ρ^{(s)} = δ_{s, t}$ . $z_{σ} = z_{σ}^{(s)} \exp (i ω_{S} t) = ρ_{σ}^{(s)} \exp (i (ϕ_{S} + ω_{S} t))$ . Hence, our solution is $η_{σ} = ρ_{σ}^{(s)} \cos (ϕ_{S} + ω_{S} t)$ . Our general solution is then, $Z_{σ} = \sum_{S} C^{(S)} ρ_{σ}^{(S)} \exp (i (ϕ_{S} ω_{S} t))$ , $H_{σ} = \sum_{S} ℜ {C^{(S)}} ρ_{σ}^{(S)} \cos (ϕ_{S} + ω_{S} t)$ .

Example

Consider the coupled pendulum, each hanging from a celing by a string of length $l$ and the (equal $m$ ) masses are connected to each other by a spring with constant $k$ and length $d_{0}$ apart. Consider a perturbation.

Let $η_{n}$ be the horizontal displacement of mass n. Then, $η_{n} / ℓ = \sin θ_{i} \approx θ_{i}$ . $1 - \cos θ \approx \frac{1}{2} θ^{2}$ . So, ${\dot{η}}_{i} = ℓ {\dot{θ}}_{i}$ .

Let $ϕ_{1}$ be the angle of the first mass and $ϕ_{2}$ be the angle of the second mass.

\begin{aligned} L & = \frac{1}{2} m ℓ^{2} ({\dot{ϕ}}_{1}^{2} + {\dot{ϕ}}_{2}^{2}) - (- m g (ℓ \cos ϕ_{1} + ℓ \cos ϕ_{2})) + \dots \\ = \frac{1}{2} m ({\dot{η}}_{1}^{2} + {\dot{η}}_{2}^{2}) - (\frac{1}{2} k (η_{1} - η_{2})^{2}) - (\frac{1}{2 ℓ} m g (η_{1}^{2} + η_{2}^{2})) \end{aligned}

Step 2: $m {\ddot{η}}_{1} + (\frac{m g}{ℓ} + k) η_{1} - k η_{2} = 0$ . $m {\ddot{η}}_{2} + (\frac{m g}{ℓ} + k) η_{2} - k η_{1} = 0$ .

Thus,

\begin{aligned} M & = (\begin{array}{c} m & 0 \\ 0 & m \end{array}) . \\ V & = (\begin{array}{c} \frac{m g}{ℓ} + k & - k \\ - k & \frac{m g}{ℓ} + k \end{array}) . \end{aligned}

Step 3: $| V - ω_{S}^{2} M | = | \begin{matrix} \frac{m g}{ℓ} + k - ω_{S}^{2} m & - k \\ - k & \frac{m g}{ℓ} + k - ω_{S}^{2} m \end{matrix} | = 0$ . So, ${(\frac{m g}{ℓ} + k - ω_{S}^{2} m)}^{2} - k^{2} = 0$ . This gives solutions, $ω_{1} = \sqrt{\frac{g}{ℓ}}, ω_{2} = \sqrt{\frac{g}{ℓ} + \frac{2 k}{m}}$ . Now, get the eigenvectors! $(\begin{matrix} \frac{m g}{ℓ} + k - \frac{g}{ℓ} m & - k \\ - k & \frac{m g}{ℓ} + k - \frac{g}{ℓ} m \end{matrix}) ρ^{(1)} k ρ_{1}^{(1)} - k ρ_{2}^{(2)} = 0$

$(\begin{matrix} \frac{m g}{ℓ} + k - (\frac{g}{ℓ} + \frac{2 k}{m}) m & - k \\ - k & \frac{m g}{ℓ} + k - (\frac{g}{ℓ} + \frac{2 k}{m}) m \end{matrix}) ρ^{(2)} = 0$ $(\begin{matrix} \frac{m g}{ℓ} + k - (\frac{g}{ℓ} + \frac{2 k}{m}) m & - k \\ - k & \frac{m g}{ℓ} + k - (\frac{g}{ℓ} + \frac{2 k}{m}) m \end{matrix}) ρ^{(2)} = - k ρ_{1}^{(2)} - k ρ_{2}^{(2)} = 0$ Thus, we get the eigenvectors, $ρ^{(1)} = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ 1 \end{matrix})$ and $ρ^{(2)} = \frac{1}{\sqrt{2}} (\begin{matrix} 1 \\ - 1 \end{matrix})$ . Normalizing with respect to the mass matrix: we get the eigenvectors, $ρ^{(1)} = \frac{1}{\sqrt{2 m}} (\begin{matrix} 1 \\ 1 \end{matrix})$ and $ρ^{(2)} = \frac{1}{\sqrt{2 m}} (\begin{matrix} 1 \\ - 1 \end{matrix})$ . The general solution is then, $η = \sum_{S} C^{(S)} ρ^{(S)} \cos (ϕ_{S} + ω_{S} t)$ .

Note that $η_{1}$ is the first row and $η_{2}$ is the second row in this vector form.

Suppose $\dot{η} (0) = 0$ and $η (0) = (\begin{matrix} α \\ 0 \end{matrix})$ … Then, $C_{1} = \frac{α \sqrt{m}}{\sqrt{2}}$ , $C_{2} = \frac{α \sqrt{m}}{\sqrt{2}}$ , $ϕ_{1} = ϕ_{2} = 0$ .

For each $ω_S4 we have $ρ^{(S)}$ a column vector. The Modal matrix, $A = (\begin{matrix} | & | & \dots & | \\ ρ^{(1)} & ρ^{(2)} & \dots & ρ^{(n)} \\ | & | & \dots & | \end{matrix})$ . Then, $A^{T} M A = I$ . But also, $A^{T} V A = (\begin{matrix} ω_{1}^{2} \\ ω_{2}^{2} \\ \dots \\ ω_{n}^{2} \end{matrix}) = Ω$ .

Normal Coordinates: $η (t) = A ξ (t)$ , $ξ (t)$ is called the normal coordinate. $A^{T} m η (t) = A^{T} m A ξ (t) = ξ (t)$ .

Recall the Lagrangian:

\begin{aligned} L & = T - V \\ = \frac{1}{2} \sum_{σ} \sum_{λ} M_{σ}^{λ} {\dot{η}}_{σ} {\dot{η}}_{λ} - V_{σ}^{λ} η_{σ} η_{λ} \\ = \frac{1}{2} {\dot{η}}^{T} M \dot{η} - \frac{1}{2} η^{T} V η \\ = \frac{1}{2} ((A \dot{ξ} (t))^{T} m (A \dot{ξ} (t)) - (A ξ (t))^{T} V (A ξ (t))) \\ = \frac{1}{2} ({\dot{ξ}}^{2} (t) - ξ^{T} (t) Ω ξ (t)) . \end{aligned}

So, we get $\frac{d}{d t} \frac{\partial L}{\partial {\dot{ξ}}_{i}} - \frac{\partial L}{\partial ξ_{i}} {\ddot{ξ}}_{i} (t) - ω_{i}^{2} ξ_{i} (t) = 0$ for each coordinate. Thus, $ξ_{σ} (t) = C^{(σ)} \cos (ω_{σ} t + ϕ_{σ})$ . We can then get, $η (t) = A ξ (t)$ .

Equation of Motion for Continuous String

Newtonian Approach

Consider a horizontal string anchored on both ends. Now, we perturb it vertically.

The mass of the string per length is $σ (x)$ .

Consider a $d x$ piece of the string. The vertical displacement at that point at a given time is going to be given by $u (x, t)$ . So, the end of the segment is $u (x + d x, t)$ . The tension is then $T_{L} = T (x)$ and $T_{R} = T (x + d x)$ . The angle at the bottom, $x$ , is $θ$ and the angle at $x + d x$ is $ϕ$ . So, $\sin θ \approx θ \approx \tan θ = \frac{\partial u (x, t)}{\partial x}$ . Similarly, $\sin ϕ \approx ϕ \approx \tan θ = \frac{\partial u (x + d x, t)}{\partial x}$ . So, $m a_{y} = [σ (x) d x] [\frac{\partial^{2} u (x, t)}{\partial t^{2}}] = T (x + d x) \sin ϕ - T (x) \sin θ = T (x + d x) \frac{\partial u (x + d x, t)}{\partial x} - T (x) \frac{\partial u (x, t)}{\partial x}$ . Let $f (x, t) = T (x) \frac{\partial u (x, t)}{\partial x}$ . Then, $[σ (x) d x] [\frac{\partial^{2} u (x, t)}{\partial t^{2}}] = f (x + d x, t) - f (x, t)$ So, $[σ (x) d x] [\frac{\partial^{2} u (x, t)}{\partial t^{2}}] = T (x) \frac{\partial u (x, t)}{\partial x} + \frac{\partial}{\partial x} (T (x) \frac{\partial u (x, t)}{\partial x}) d x + \dots - T (x) \frac{\partial u (x, t)}{\partial x} = \frac{\partial}{\partial x} (T (x) \frac{\partial u (x, t)}{\partial x}) d x$ . So, $σ (x) \frac{\partial^{2} u (x, t)}{\partial t^{2}} = \frac{\partial}{\partial x} (T (x) \frac{\partial u (x, t)}{\partial x})$ . Hence, $σ (x) \frac{\partial^{2} u (x, t)}{\partial t^{2}} = T (x) \frac{\partial^{2} u (x, t)}{\partial x^{2}} + T^{'} (x) \frac{\partial u (x, t)}{\partial x}$ . For mass density and tension position independent, $\frac{\partial^{2} u (x, t)}{\partial t^{2}} = \frac{T}{σ} \frac{\partial^{2} u (x, t)}{\partial x^{2}}$ which can be rewritten as $(\frac{1}{c^{2}} \frac{\partial^{2}}{\partial t^{2}} - \frac{\partial^{2}}{\partial x^{2}}) u (x, t) = 0$ .

Hamilton’s Principle in a Continuous Medium

Consider breaking it up into $N$ masses–then take the limit to $N \to \infty$ . Let the position of the $n-$th mass be $x_{n}$ and the vertical displacement $u_{n}$ . Consider the spacing between the masses as $a$ .

Writing the lagrangian, with a tension of $τ$ , so $k = \frac{τ}{a}$ .

\begin{aligned} L & = \sum_{i = 1}^{N} \frac{1}{2} m_{i} {\dot{u}}_{i}^{2} - \sum_{i = 0}^{N} \frac{1}{2} k (u_{i + 1} - u_{i})^{2} \\ = \sum_{i = 1}^{N} \frac{1}{2} \frac{m_{i}}{a} {\dot{u}}_{i}^{2} a - \sum_{i = 0}^{N} \frac{1}{2} τ {(\frac{u_{i + 1} - u_{i}}{a})}^{2} a . \end{aligned}

Let $N \to \infty$ . So, $a \to d x$ , $\frac{m}{a} = σ (x)$ , $u_{i} (x, t) \to u (x, t)$ .

\begin{aligned} L & = \int_{0}^{ℓ} \frac{1}{2} σ (x) {(\frac{\partial u (x)}{\partial t})}^{2} d x - \int_{0}^{ℓ} \frac{1}{2} τ (x) {(\frac{\partial u (x, t)}{\partial x})}^{2} d x \\ = \frac{1}{2} \int_{0}^{ℓ} σ (x) {(\frac{\partial u (x)}{\partial t})}^{2} - τ (x) {(\frac{\partial u (x, t)}{\partial x})}^{2} d x . \end{aligned}

Then, the lagrangian is an integral of the lagrangian density,

\begin{aligned} L & = \int_{0}^{ℓ} L (u (x, t), \frac{\partial u (x, t)}{\partial x}, \frac{\partial u (x, t)}{\partial t}, x, t) d x . \end{aligned}

So,

\begin{aligned} I & = \int_{t_{1}}^{t_{2}} L d t \\ = \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} L (u (x, t), \frac{\partial u (x, t)}{\partial x}, \frac{\partial u (x, t)}{\partial t}, x, t) d x d t . \\ δ I & = 0. \end{aligned}

Notice, $δ u (x, t_{1}) = δ u (x, t_{2}) = 0 = δ u (0, t) = δ u (ℓ, t)$ .

\begin{aligned} δ I & = δ \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} L (u (x, t), \frac{\partial u (x, t)}{\partial x}, \frac{\partial u (x, t)}{\partial t}, x, t) d x d t \\ = \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} δ L (u (x, t), \frac{\partial u (x, t)}{\partial x}, \frac{\partial u (x, t)}{\partial t}, x, t) d x d t \\ = \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} \frac{\partial L}{\partial u} δ u + \frac{\partial L}{\partial u^{'}} \frac{\partial}{\partial x} δ u + \frac{\partial L}{\partial \dot{u}} \frac{\partial}{\partial t} δ u d x d t \\ = \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} (\frac{\partial L}{\partial u} + \frac{\partial L}{\partial u^{'}} \frac{\partial}{\partial x} + \frac{\partial L}{\partial \dot{u}} \frac{\partial}{\partial t}) δ u d x d t \\ = \dots (I B P) \\ = \int_{t_{1}}^{t_{2}} \int_{0}^{ℓ} (\frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \frac{\partial L}{\partial u^{'}} - \frac{\partial}{\partial t} \frac{\partial L}{\partial \dot{u}}) δ u d x d t \\ 0 & = (\frac{\partial L}{\partial u} - \frac{\partial}{\partial x} \frac{\partial L}{\partial u^{'}} - \frac{\partial}{\partial t} \frac{\partial L}{\partial \dot{u}}) \end{aligned}

Note that I abbreviated $u^{'} = \frac{\partial u}{\partial x}$ and $\dot{u} = \frac{\partial u}{\partial t}$ . Recall,

\begin{aligned} L & = σ (x) {(\frac{\partial u (x)}{\partial t})}^{2} - τ (x) {(\frac{\partial u (x, t)}{\partial x})}^{2} \end{aligned}

If $σ (x) = σ$ and $τ (x) = τ$ . Then,

\begin{aligned} 0 & = 0 - (- 2 τ \frac{\partial^{2} u (x, t)}{\partial x^{2}}) - (2 σ \frac{\partial^{2} u (x, t)}{\partial t^{2}}), \\ (τ \frac{\partial^{2} u (x, t)}{\partial x^{2}}) & = (σ \frac{\partial^{2} u (x, t)}{\partial t^{2}}) . \end{aligned}

Giving us our wave equation again.