The Pendulum#

The ideal simple pendulum – a massless rigid rod of length \(l\) with a point mass \(m\) fixed to move about a pivot – is a familiar and somewhat intuitive physical systems that contains within it a wealth of physics.

Intuitive Questions#

You should be able to justify these intuitive questions using the formal results derived below. Use the formalism to test and check your intuition. For all of these questions, be prepared to explain you answer both quantitatively using the formal results, and intuitively.

How does the oscillation frequency of a pendulum depend on its amplitude?
Your frisbee is stuck in a tall narrow tree. It is fairly precariously balanced, so you are sure you can recover it if you shake the tree hard enough. You give the tree a nudge and notice the natural oscillation frequency. Should you vibrate the tree at a slightly higher frequency or a slightly lower frequency to maximize the chance of recovering your frisbee?
You have a tall vase on your table. If you vibrate the table at the resonant frequency of the vase, it might fall over. Do you need to be concerned about vibrating the table at any other frequency, like twice or half of the resonant frequency?
A pendulum with very little friction sits at equilibrium. You can vibrate the pendulum at a fixed frequency of your choice (with low amplitude), but you cannot change the frequency once you choose it. Can you get the pendulum to complete a full revolution (swing over the top)?
Describe how the amplitude of an oscillating pendulum with low damping changes if you slowly change the length of the pendulum. (E.g., consider a mass oscillating at the end of a long rope as you pull the rope up through a hole.) What about if you have a pendulum of fixed length in an elevator that gradually changes acceleration? (Is this the same problem, or is there a difference?)
Consider balancing a ruler standing on its end. You probably know that you can balance it if you allow your hand to move laterally - i.e. chasing the ruler as it starts to fall. Can you somehow balance the ruler by only moving your hand up and down? (You can clasp the ruler so that you can push it up and pull it down, but your fingers do not have enough strength to apply significant torque to the ruler if you cannot move your hand laterally.)

Model#

The complete model we shall consider is

\[\begin{gather*} m\ddot{\theta} = -\frac{mg}{l}\omega^2\sin\theta - 2m\lambda\dot{\theta} + \Gamma \end{gather*}\]

where all quantities might possibly depend on time:

\([\theta] = 1\): Angle of the pendulum from equilibrium (hanging down). This is the dynamical variable. When we generalize the problem, or discuss the small amplitude limit, we shall use the general coordinate notation \(q=\theta\).
\([m] = M\): Mass of the pendulum bob. As there is only one dynamical mass in this problem, we will remove it below.
\([g] = D/T^2\): Acceleration due to gravity, also known as the gravitational field.
\([l] = D\): Length of the pendulum.
\([\lambda] = 1/T\): Damping, either due to friction at high speed, wind resistance, or dragging an object through a viscous fluid. Note: this is not standard \(F_f = \mu F_N\) friction (which makes another interesting problem).
\([\Gamma] = M/T^2\): Driving torque (\(F = l\Gamma\) if the force is directed perpendicular to the pendulum rod).

Canceling the common factor of mass and write this as:

\[\begin{gather*} \ddot{\theta} + 2\lambda\dot{\theta} + \omega^2_0(t)\sin\theta = f(t), \qquad \omega^2_0 = \frac{g}{l}, \qquad f = \frac{\Gamma}{m}. \end{gather*}\]

\([\omega_0] = 1/T\): Only the combination \(g/l\) has physical significance for this system if the driving force is appropriately defined. This is sometimes called the natural resonant frequency of the system. It is the angular frequency in the small amplitude limit with no damping: i.e. the harmonic oscillator. Allowing the frequency to change with time \(\omega_0(t)\) allows us to study adiabatic invariants (slow variations), parametric resonance (small oscillations over a range of frequencies), and Kapitz’s pendulum (rapid small oscillations).
\([f] = 1/T^2\): Driving force/torque expressed as an acceleration, torque per unit mass, or force per unit mass and unit length.

With these re-definitions, everything is specified in terms of time/frequency. One might further define \(\omega_0=1\) or similar rendering the problem dimensionless, but we will not do so here as it obscures some of the physics.

Many more details will be discussed in Worked Example: The Pendulum, but provide here a set of problems for you to work through to test your understanding of mechanics. I have tried to express these simply so that you can try to attack them without knowing specific techniques like the Hamilton-Jacobi equation. My hope is that, after trying these and making mistakes, you will appreciate better the power of some of the more formal techniques of classical mechanics.

Harmonic Oscillator#

Here we consider small amplitude oscillations \(\theta \ll 1\), keeping only the first term in

\[\begin{gather*} \sin\theta = \theta - \frac{\theta^3}{3!} + O(\theta^5). \end{gather*}\]

Do It!

Find the general solution for a harmonic oscillator:

\[\begin{gather*} \ddot{q} + \omega^2 q = 0, \qquad q(0) = q_0, \qquad \dot{q}(0) = \dot{q}_0. \end{gather*}\]

Solution

\[\begin{gather*} q(t) = q_0\cos\omega t + \frac{\dot{q}_0}{\omega}\sin \omega t. \end{gather*}\]

Express the Hamiltonian and Lagrangian formulations.

\[\begin{gather*} T = \frac{\dot{q}^2}{2}, \qquad V = \frac{\omega^2q^2}{2}, \qquad L(q, \dot{q}, t) = T - V = \frac{\dot{q}^2}{2} - \frac{\omega^2q^2}{2}\\ p = L_{,\dot{q}} = \dot{q}, \qquad H = p\dot{q} - L = \frac{p^2}{2} + \frac{\omega^2q^2}{2}. \end{gather*}\]

Damped Harmonic Oscillator#

Make sure you understand how to solve general homogeneous linear ODEs like this one.

Do It!

Find the general solution for a damped harmonic oscillator:

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega^2 q = 0, \qquad q(0) = q_0, \qquad \dot{q}(0) = \dot{q}_0. \end{gather*}\]

Solution

\[\begin{gather*} q(t) = e^{-\lambda t}\left( q_0\cos\bar{\omega} t + \frac{\dot{q}_0 + \lambda q_0}{\bar{\omega}}\sin \bar{\omega} t \right), \qquad \bar{\omega} = \sqrt{\omega^2 - \lambda^2}. \end{gather*}\]

Express the Hamiltonian and Lagrangian formulations.

\[\begin{gather*} T = \frac{\dot{q}^2}{2}, \qquad V = \frac{\omega^2q^2}{2}, \qquad L(q, \dot{q}, t) = e^{2\lambda t} \left(\frac{\dot{q}^2}{2} - \frac{\omega^2q^2}{2}\right)\\ p = L_{,\dot{q}} = e^{2\lambda t}\dot{q}, \qquad H = p\dot{q} - L = e^{-2\lambda t}\frac{p^2}{2} + e^{2\lambda t}\frac{\omega^2q^2}{2}. \end{gather*}\]

This is probably not obvious. Check explicitly that you get the correct equation of motion.

\[\begin{gather*} \overbrace{2\lambda e^{2\lambda t}\dot{q} + e^{2\lambda t}\ddot{q}}^{\dot{p}} = \overbrace{-e^{2\lambda t}\omega^2q}^{L_{,q}},\\ 2\lambda \dot{q} + \ddot{q} = -\omega^2q. \end{gather*}\]

Driven Damped Harmonic Oscillator#

Do It!

Find the general solution for a damped harmonic oscillator:

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega^2 q = f(t). \end{gather*}\]

Make sure you have a general solution, but don’t worry about expressing the coefficients. Your answer should be expressed in terms of various integrals.

Solution

The general solution can be written as

\[\begin{multline*} q(t) = e^{-\lambda t}\cos \bar{\omega} t \int \frac{-f(t)e^{-\lambda t}\sin\bar{\omega}t}{\bar{\omega}}\;\d{t}\\ + e^{-\lambda t}\sin \bar{\omega} t \int \frac{f(t)e^{-\lambda t}\cos\bar{\omega}t}{\bar{\omega}}\;\d{t}. \end{multline*}\]

The constants of integration provide the general solution.

This follows from the fact that, given two independent solutions \(A(t)\) and \(B(t)\) of a homogeneous linear differential equation, the solution \(q(t)\) to an inhomogeneous equation can be formed as follows:

\[\begin{gather*} \ddot{A} + g_1(t)\dot{A} + g_0(t) A = 0,\\ \ddot{B} + g_1(t)\dot{B} + g_0(t) B = 0,\\ \ddot{q} + g_1(t)\dot{q} + g_0(t) q = f(t), \end{gather*}\]

\[\begin{gather*} q = aA + bB,\\ \dot{a} = -\frac{fB}{W},\qquad \dot{b} = \frac{fA}{W},\\ W = A\dot{B} - B\dot{A}. \end{gather*}\]

To check, just compute:

\[\begin{gather*} q = aA + bB,\\ \dot{q} = \underbrace{\dot{a}A + \dot{b}B}_{0} + a\dot{A} + b\dot{B} = a\dot{A} + b\dot{B},\\ \ddot{q} = \underbrace{\dot{a}\dot{A} + \dot{b}\dot{B}}_{f} + a \ddot{A} + b\ddot{A} = f + a \ddot{B} + b\ddot{A}. \end{gather*}\]

The key is to notice that the remaining terms when inserted into the original equation, simply fulfill the homogeneous equation, with the inhomogeneous term leftover.

\[\begin{multline*} \ddot{q} + g_1(t)\dot{q} + g_0(t) q = a\overbrace{(\ddot{A} + g_1(t)\dot{A} + g_0(t)A)}^{0} \\ + b\underbrace{(\ddot{B} + g_1(t)\dot{B} + g_0(t)B)}_{0} + f(t) = f(t). \end{multline*}\]

The key is to choose the coefficients in each step to eliminate the derivatives of the coefficients, and in the last step, reproduce the inhomogeneous term. For a second-order differential equation, this gives the following two-dimensional linear system, which can be solved by Cramer’s rule

\[\begin{gather*} \begin{pmatrix} A & B\\ \dot{A} & \dot{B} \end{pmatrix} \begin{pmatrix} \dot{a}\\ \dot{b} \end{pmatrix} = \begin{pmatrix} 0\\ f \end{pmatrix}, \\ \begin{pmatrix} \dot{a}\\ \dot{b} \end{pmatrix} = \begin{pmatrix} A & B\\ \dot{A} & \dot{B} \end{pmatrix}^{-1} \begin{pmatrix} 0\\ f \end{pmatrix} = \frac{1}{A\dot{B} - B\dot{A}} \begin{pmatrix} \dot{B} & -B\\ -\dot{A} & A \end{pmatrix} \begin{pmatrix} 0\\ f \end{pmatrix}. \end{gather*}\]

This method easily generalizes to higher order.

Do It!

Compute the linear response \(\chi\) of the damped harmonic oscillator:

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega_0^2 q = f \cos \omega t. \end{gather*}\]

After a time long enough for any transients to die off, the system will simply oscillate with some amplitude \(a\) at the drive frequency \(\omega\) but with a phase offset \(\phi\) form the driving force:

\[\begin{gather*} q(t) = a\cos(\omega t + \phi). \end{gather*}\]

The density linear response is the complex quantity

\[\begin{gather*} \chi = \frac{a}{f}e^{\I\phi}. \end{gather*}\]

Plot \(\abs{\chi}^2\) as a function of \(\omega\). At what angular frequency \(\bar{\omega}\) is this a maximum? What is the phase shift \(\phi\) at this maximum?

Solution

This problem is most easily solved using Fourier techniques. The equations of motion can be expressed as the real part of

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega_0^2 q = f e^{\I \omega t}, \qquad q(t) = \chi f e^{\I\omega t} = a e^{\I \omega t + \phi},\\ (-\omega^2 + 2 \I\omega \lambda + \omega_0^2)\chi fe^{\I\omega t} = fe^{\I\omega t},\\ \chi = \frac{1}{\omega_0^2 - \omega^2 + 2 \I \omega \lambda} = \frac{\omega_0^2 - \omega^2 - 2 \I\omega \lambda} {(\omega_0^2 - \omega^2)^2 + 4\omega^2\lambda^2},\\ \abs{\chi}^2 = \frac{1}{(\omega_0^2 - \omega^2)^2 + 4\omega^2\lambda^2}, \qquad \tan\phi = \frac{2\omega\lambda}{\omega^2 - \omega_0^2}. \end{gather*}\]

Completing the square, we have

\[\begin{gather*} \abs{\chi}^2 = \frac{1} {\bigl(\omega^2 - (\omega_0^2 - 2\lambda^2)\bigr)^2 + 4\lambda^2(\omega_0^2 - \lambda^2)}. \end{gather*}\]

Note that the resonance peak occurs at the shifted frequency

\[\begin{gather*} \omega_{\max}^2 = \omega_0^2 - 2\lambda^2 \leq \omega_0^2,\qquad \abs{\chi_{\max}}^2 = \frac{1} {4\lambda^2(\omega_0^2 - \lambda^2)}. \end{gather*}\]

which is not the same as the natural frequency \(\bar{\omega}^2 = \omega_0^2 - \lambda^2\) of the damped oscillator. At resonance, the phase shift is

\[\begin{gather*} \tan\phi = -\sqrt{\frac{\omega_0^2}{\lambda^2} - 2}. \end{gather*}\]

Note that if the damping is small \(\lambda \ll \omega_0\), then

\[\begin{gather*} \tan\phi \approx -\frac{\omega_0}{\lambda}, \qquad \phi \approx -\frac{\pi}{2} + \frac{\lambda}{\omega_0} + O\left(\frac{\lambda^2}{\omega_0^2}\right). \end{gather*}\]

Show code cell content Hide code cell content

w0 = 1.0
w = np.linspace(0, 2*w0, 500)
lams = [0, 0.1, 0.2, 0.3]

fig, ax = plt.subplots(figsize=(4, 3))
for lam in lams:
    chi = 1/(w0**2 - w**2 + 2j*w*lam)
    l, = ax.semilogy(w/w0, abs(chi*w0**2)**2, label=f"$\lambda={lam:.2g}\omega_0$")
    if lam > 0:
        w_max = np.sqrt(w0**2 - 2*lam**2)
        chi2_max = 1/(4*lam**2*(w0**2-lam**2))
        ax.plot([w_max/w0], [chi2_max*w0**4], 'o', c=l.get_c())

lams = np.linspace(0, w0, 100)[1:-1]
w_max = np.sqrt(w0**2 - 2*lams**2)
chi2_max = 1/(4*lams**2*(w0**2-lams**2))
ax.plot(w_max/w0, chi2_max*w0**4, 'k:')
ax.legend()
ax.set(xlabel=r"$\omega/\omega_0$",
       ylabel=r"$|\chi|^2\omega_0^4$",
       ylim=(0, 40))

if glue: glue("fig:LinearResponse", fig);
plt.tight_layout()
fig.savefig(FIG_DIR / "LinearResponse.svg")

<>:8: SyntaxWarning: invalid escape sequence '\l'
<>:8: SyntaxWarning: invalid escape sequence '\o'
<>:8: SyntaxWarning: invalid escape sequence '\l'
<>:8: SyntaxWarning: invalid escape sequence '\o'
/tmp/ipykernel_5554/1627739606.py:8: SyntaxWarning: invalid escape sequence '\l'
  l, = ax.semilogy(w/w0, abs(chi*w0**2)**2, label=f"$\lambda={lam:.2g}\omega_0$")
/tmp/ipykernel_5554/1627739606.py:8: SyntaxWarning: invalid escape sequence '\o'
  l, = ax.semilogy(w/w0, abs(chi*w0**2)**2, label=f"$\lambda={lam:.2g}\omega_0$")
/tmp/ipykernel_5554/1627739606.py:8: SyntaxWarning: invalid escape sequence '\l'
  l, = ax.semilogy(w/w0, abs(chi*w0**2)**2, label=f"$\lambda={lam:.2g}\omega_0$")
/tmp/ipykernel_5554/1627739606.py:8: SyntaxWarning: invalid escape sequence '\o'
  l, = ax.semilogy(w/w0, abs(chi*w0**2)**2, label=f"$\lambda={lam:.2g}\omega_0$")

---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
Cell In[2], line 2
      1 w0 = 1.0
----> 2 w = np.linspace(0, 2*w0, 500)
      3 lams = [0, 0.1, 0.2, 0.3]
      5 fig, ax = plt.subplots(figsize=(4, 3))

NameError: name 'np' is not defined

Parametric Resonance#

Another way of driving a system is to vary a parameter, for example,

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega^2(t)q = 0, \qquad \omega(t+T) = \omega(t), \qquad \omega_0 = \frac{2\pi}{T}. \end{gather*}\]

The key feature of a parametric resonance is that they required a perturbation to get started. If we start from the equilibrium position \(q = 0\), then – even with the drive – we stay there. Some other features of parametric resonances:

Unlike with driven systems where there is only one resonant frequency \(\omega_0\), parametric resonances occur at a set of frequencies:

\[\begin{gather*} \omega_n = 2\frac{\omega_0}{n}. \end{gather*}\]
Parametric resonances tend to be very narrow, especially for large \(n\).
They also quite sensitive to damping, and require a finite-size amplitude \(\epsilon > \epsilon_n\) in the presence of damping.

Do It!

Consider \(\omega^(t) = \omega_0^2(1 + \epsilon \sin \omega t)\). Try to find a parametric resonance numerically.

Anharmonic Oscillator#

En-route to a pendulum, consider adding an anharmonic perturbation:

\[\begin{gather*} \ddot{q} + 2\lambda \dot{q} + \omega_0^2(q + \epsilon q^3) = 0. \end{gather*}\]

Do It!

Treating \(\epsilon \ll 1\) as a small parameter, use some sort of perturbation theory to estimate how this perturbation changes the natural frequency of the oscillations to lowest order in \(\epsilon\). Does your calculation support your intuition? If you were to compute higher order corrections (i.e. express \(\omega\) as a power series in \(\epsilon\)), what would you expect the radius of convergence in \(\epsilon\) to be?

Solution

First some qualitative discussion. The corresponding potential is

\[\begin{gather*} V(q) = \omega_0^2\frac{q^2}{2} + \epsilon \omega_0^2 \frac{q^4}{4}. \end{gather*}\]

This becomes more and more confining as the amplitude grows, so I would expect that the frequency should increase with increasing amplitude.

Suppose that we could compute the power-series for \(\omega = \omega_0 + \epsilon \omega_1 + \epsilon^2 \omega_2 + \cdots\) where \(\omega_{n\geq 1}(E)\) depend on the amplitude of the oscillations. If this converges, it should converge in a region of the complex plane with \(\abs{\epsilon} < R_\epsilon\) where \(R_\epsilon\) is the radius of convergence. However, if \(\epsilon < 0\), then the potential is unbounded below, and once oscillations become sufficiently large, there likely ceases to be a reasonable frequency. This might indicate that the radius of converges is zero (i.e. the series is an divergent asymptotic series.)

Starting with the case of \(\lambda = 0\) and applying naïve perturbation theory to the solution with initial conditions \(q(0) = q_0\) and \(\dot{q}(0) = 0\) gives the solution (cf. [Fetter and Walecka, 2006] (7.29))

\[\begin{gather*} q(t) = q_0\cos \omega_0 t + \epsilon\omega_0^2 \frac{q_0^3}{32}( \cos 3\omega_0 t - \cos\omega_0 t - 12 \omega t \sin \omega_0 t) + O(\epsilon^2). \end{gather*}\]

Note that this solution still two fundamental issues:

It still has a frequency \(\omega_0\) (although now it has some harmonics). This means that as time progresses, the errors will get larger and larger as the approximation gets out of phase with the full solution.
Worse, there is a term proportional with \(t\) that grows linearly and in an unbounded manner! Expanding for small \(t\), we have

\[\begin{gather*} q(t) = q_0 - \frac{\omega_0^2 t^2}{2} + \epsilon\omega_0^4 \frac{q_0^3}{2}t^2 + O(\epsilon^2) \end{gather*}\]

suggesting that

\[\begin{gather*} \omega^2 \approx \omega_0^2\left(1 - \epsilon \omega_0^2 q_0^3\right). \end{gather*}\]

Thus, even if we interpret the small short time behavior as indicative of how the period will change, we get the wrong impression that the frequency should decrease.

These issues indicate serious deficiencies with classical perturbation theory. We discuss them and their resolution in Perturbation Theory.

Full Pendulum#

We finally consider the full pendulum:

\[\begin{gather*} \ddot{q} + 2\lambda\dot{q} + \omega^2(t)q = f(t). \end{gather*}\]

Kapitza’s Pendulum#

What happens if we consider a parametric oscillator (\(f(t)=0\)) driven at a high frequency?

\[\begin{gather*} \omega^2(t) \omega_0^2(1 + \epsilon \sin \omega t), \qquad \omega \gg \omega_0. \end{gather*}\]

This problem was considered by Pyotr Kaptitza, who showed that one can use such a mechanism to stabilize the motion about unstable points.

Do It!

The essence of the solution is to describe the solution in terms of two time-scales:

\[\begin{gather*} q(t) \approx q_0(t) + \Re A(t)e^{\I\omega t} \end{gather*}\]

where the \(q_0(t)\) and \(A(t)\) vary slowly. Their behavior can be estimated by averaging over the high frequency oscillations. See if you can work out the effective equations for this motion. (A full solution to this problem is provided in [Landau and Lifshitz, 1976] if you need inspiration, and a sophisticated general analysis is provided in [Arnol'd, 1989].)

For a full solution, see Kapitza’s Pendulum.

The Pendulum

Contents

The Pendulum#

Intuitive Questions#

Model#

Harmonic Oscillator#

Damped Harmonic Oscillator#

Driven Damped Harmonic Oscillator#

Parametric Resonance#

Anharmonic Oscillator#

Full Pendulum#

Kapitza’s Pendulum#