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 \(t=0\).