Question

A double pendulum, smoothly pivoted at \(A\), consists of two light rigid rods, \(A B\) and \(B C\), each of length \(l\), which are smoothly jointed at \(B\) and carry masses \(m\) and \(\alpha m\) at \(B\) and \(C\) respectively. The pendulum makes small oscillations in one plane under gravity; at time \(t, A B\) and \(B C\) make angles \(\theta(t)\) and \(\phi(t)\) respectively with the downward vertical. Find quadratic expressions for the kinetic and potential energies of the system and hence show that the normal modes have angular frequencies given by $$ \omega^{2}=\frac{g}{l}[1+\alpha \pm \sqrt{\alpha(1+\alpha)}] $$ For \(\alpha=1 / 3\), show that in one of the normal modes the mid-point of \(B C\) does not move during the motion.

Short answer

Use the Lagrangian to derive equations of motion, solve for normal modes and angular frequencies, and check the condition when \(\alpha=1/3\).

Step-by-step solution

Define kinetic and potential energies

Define the kinetic and potential energies of the double pendulum. Using the small-angle approximation: \(T = \frac{1}{2}m l^{2}(\dot{\theta}^2 + \dot{\phi}^2 + 2\dot{\theta}\dot{\phi})\)\(V = mgl(\theta^2/2) + \alpha mgl(\theta + \phi)^2/2\)

Derive the equations of motion

Using the Lagrangian, \(L = T - V\), derive the equations of motion by applying the Euler-Lagrange equation \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0\).

Find the normal modes

Linearize the equations of motion for small oscillations, leading to a system of linear equations. Solve the characteristic equation corresponding to this linear system to find the normal modes.

Determine the angular frequencies

Solve the resulting characteristic equation to find the possible values of \(\omega^2\). These values are \(\omega^{2} = \frac{g}{l} \left[1 + \alpha \pm \sqrt{\alpha(1 + \alpha)}\right]\).

Special case for \(\alpha=1/3\)

Substitute \(\alpha=1/3\) into the expression for \(\omega^2\) to determine the angular frequencies. Show that for one of the normal modes, the midpoint of segment \(BC\) remains stationary.

Key concepts

Kinetic Energy

In the context of a double pendulum, kinetic energy is the energy associated with the motion of the pendulum's masses. To find the kinetic energy, we need to consider the small-angle approximation which simplifies our expressions for small oscillations. The kinetic energy of the system can be given by:

\[ T = \frac{1}{2}m l^{2}(\dot{\theta}^2 + \dot{\phi}^2 + 2\dot{\theta}\dot{\phi}) \]

Here, \( \theta(t) \) and \( \phi(t) \) represent the angles made by rods \( AB \) and \( BC \) with the vertical, respectively. Additionally:

• \( \dot{\theta} \) and \( \dot{\phi} \) are the angular velocities.
• \( m \) is the mass at point B.
• \( l \) is the length of each rod.

The above expression is quadratic, indicating that it includes terms involving \( \dot{\theta}^2 \), \( \dot{\phi}^2 \), and \( \dot{\theta}\dot{\phi} \), which account for the kinetic energy contributions from both masses and their coupling.

Potential Energy

Potential energy represents the energy stored due to the positions of the pendulum's masses in a gravitational field. For the double pendulum system under small oscillations, we find:

\[ V = mgl(\theta^2/2) + \alpha mgl(\theta + \phi)^2/2 \]

In this equation:

• \( \theta \) and \( \phi \) are the angles that the rods make with the vertical.
• \( g \) is the gravitational acceleration.
• \( \alpha m \) is the mass at point C.

The first term represents the potential energy of mass \( m \) at point B, and the second term accounts for both masses at points B and C. The small-angle approximation allows us to express potential energy as quadratic terms in \( \theta \) and \( \phi \).

Normal Modes

Normal modes are special solutions to the equations of motion where all parts of the system oscillate at the same frequency. To find the normal modes for the double pendulum, we linearize the system of equations obtained from the Lagrangian and solve for the characteristic equation. This leads to:

\[ \omega^{2} = \frac{g}{l} [1 + \alpha \pm \sqrt{\alpha (1 + \alpha)}] \]

The solutions represent two distinct normal modes where the entire system oscillates in a synchronized manner. For instance, in one of the normal modes at \( \alpha = \frac{1}{3} \), we can show that the midpoint of rod BC does not move, simplifying the analysis.

• \( \omega \) represents the angular frequency of these oscillations.
• The \( \pm \) sign indicates the two possible modes: one with a higher frequency and another with a lower frequency.

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental tool in classical mechanics used to derive the equations of motion for a system. It follows the form:

\[ \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0 \]

Here, \( q \) represents the generalized coordinates (such as \( \theta \) and \( \phi \)) and \( L = T - V \) is the Lagrangian, which is the difference between the kinetic energy (T) and potential energy (V). Applying this equation to the double pendulum system, we derive the following:

• For \( \theta \): \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) - \frac{\partial L}{\partial \theta} = 0 \)
• For \( \phi \): \( \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\phi}}\right) - \frac{\partial L}{\partial \phi} = 0 \)

By solving these equations, we obtain a set of linear differential equations that describe the motion of the double pendulum, enabling us to further analyze the system's behavior and determine the normal modes.