60.1: Examples

Example $\PageIndex{1}$ Simple Harmonic Oscillator

As an example of the use of Lagrange's equation, consider a one-dimensional simple harmonic oscillator. We wish to find the position $x$ of the oscillator at any time \(t\

Solution

We begin by writing the usual expression for the kinetic energy $K$ :

$$K=\frac{1}{2} m v^{2}$$

The potential energy $U$ of a simple harmonic oscillator is given by

$$U=\frac{1}{2} k x^{2}$$

The Lagrangian in this case is then

$$\begin{aligned} L(x, v) & =K-U \\ & =\frac{1}{2} m v^{2}-\frac{1}{2} k x^{2} \end{aligned}$$

Lagrange's equation in one dimension is

$$\frac{d}{d t}\left(\frac{\partial L}{\partial v}\right)-\frac{\partial L}{\partial x}=0$$

Substituting for $L$ from Eq. $\PageIndex{4}$, we find

$$\frac{d}{d t}\left[\frac{\partial}{\partial v}\left(\frac{1}{2} m v^{2}-\frac{1}{2} k x^{2}\right)\right]-\frac{\partial}{\partial x}\left(\frac{1}{2} m v^{2}-\frac{1}{2} k x^{2}\right)=0$$

Evaluating the partial derivatives, we get

$$\frac{d}{d t}(m v)+k x=0$$

or, since $v=d x / d t$,

$$m \frac{d^{2} x}{d t^{2}}+k x=0$$

which is a second-order ordinary differential equation that one can solve for $x(t)$. Note that the first term on the left is $m a=F$, so this equation is equivalent to $F=-k x$ (Hooke's Law). The solution to the differential equation (57.10) turns out to be

$$x(t)=A \cos (\omega t+\delta)$$

where $A$ is the amplitude of the motion, $\omega=\sqrt{k / m}$ is the angular frequency of the oscillator, and $\delta$ is a phase constant that depends on where the oscillator is at $t=0$.

Example $\PageIndex{2}$ Plane Pendulum

Part of the power of the Lagrangian formulation of mechanics is that one may define any coordinates that are convenient for solving the problem; those coordinates and their corresponding velocities are then used in place of $x$ and $v$ in Lagrange's equation.

For example, consider a simple plane pendulum of length $\ell$ with a bob of mass $m$, where the pendulum makes an angle $\theta$ with the vertical. The goal is to find the angle $\theta$ at any time $t$.

Solution

In this case we replace $x$ with the angle $\theta$, and we replace $v$ with the pendulum's angular velocity $\omega$. The kinetic energy $K$ of the pendulum is the rotational kinetic energy

$$K=\frac{1}{2} I \omega^{2}=\frac{1}{2} m \ell^{2} \omega^{2},$$

where $I$ is the moment of inertia of the pendulum, $I=m \ell^{2}$. The potential energy of the pendulum is the gravitational potential energy

$$U=m g \ell(1-\cos \theta)$$

The Lagrangian in this case is then

$$\begin{aligned} L(\theta, \omega) & =K-U \\ & =\frac{1}{2} m \ell^{2} \omega^{2}-m g \ell(1-\cos \theta) \end{aligned}$$

Lagrange's equation becomes

$$\frac{d}{d t}\left(\frac{\partial L}{\partial \omega}\right)-\frac{\partial L}{\partial \theta}=0$$

Substituting for $L$,

$$\frac{d}{d t}\left\{\frac{\partial}{\partial \omega}\left[\frac{1}{2} m \ell^{2} \omega^{2}-m g \ell(1-\cos \theta)\right]\right\}-\frac{\partial}{\partial \theta}\left[\frac{1}{2} m \ell^{2} \omega^{2}-m g \ell(1-\cos \theta)\right]=0$$

Computing the partial derivatives, we find

$$\frac{d}{d t}\left(m \ell^{2} \omega\right)+m g \ell \sin \theta=0$$

Since $\omega=d \theta / d t$, this gives

$$m \ell^{2} \frac{d^{2} \theta}{d t^{2}}+m g \ell \sin \theta=0,$$

which is a second-order ordinary differential equation that one may solve for the motion $\theta(t)$. The first term on the left-hand side is the torque $\tau$ on the pendulum, so this equation is equivalent to $\tau=-m g \ell \sin \theta$.

The solution to the differential equation ( $\PageIndex{17}$ ) is quite complicated, but we can simplify it if the pendulum only makes small oscillations. In that case, we can approximate $\sin \theta \approx \theta$, and the differential equation ( $\PageIndex{17}$ becomes a simple harmonic oscillator equation with solution

$$\theta(t) \approx \theta_{0} \cos (\omega t+\delta)$$

where $\theta_{0}$ is the (angular) amplitude of the pendulum, $\omega=\sqrt{g / \ell}$ is the angular frequency, and $\delta$ is a phase constant that depends on where the pendulum is at $t=0$.