Question

${}^{†}\mathrm{A}$ double pendulum consists of a bob of mass $m$ suspended under gravity from a fixed point by a light string of length $a$, and a second bob, also of mass $m$, suspended from the first by a further light string, also of length $a$. Show that if the motion is near equilibrium and is confined to a vertical plane through the fixed point then the normal frequencies are $\sqrt{(2\pm \sqrt{2})g/a}$. Describe the corresponding normal modes. Without doing further calculation, describe the normal modes when the motion is not confined to a vertical plane.

Short answer

In summary, for small displacements in a vertical plane, there are two normal modes of oscillation for a double pendulum system with equal mass and length. The first normal mode has both bobs oscillating in phase, moving in the same direction simultaneously with equal amplitudes. The second normal mode has the bobs oscillating out of phase, moving in opposite directions with the second bob having a higher amplitude than the first one. The normal frequencies of these oscillations are given by ${\omega }_{1,2}=\sqrt{(2\pm \sqrt{2})g/a}$. When the motion is not confined to a vertical plane, the normal modes remain in-phase and out-of-phase oscillations, but the bobs are free to oscillate in different planes in a circular motion.

Step-by-step solution

Define the coordinates and displacements

Let's denote the angles that the strings make with the vertical direction as ${\theta }_1$ for the first bob and ${\theta }_2$ for the second bob. We define $\alpha ={\theta }_1$ and $\beta ={\theta }_2-{\theta }_1$ to represent the displacements of the two bobs from their equilibrium positions.

Write down the equations for centripetal and gravitational forces

The total potential energy of the system can be written as: $V=-mg(a\cos (\alpha )+a\cos (\alpha +\beta ))$. The total kinetic energy of the system is given by: $T=\frac{1}{2}m{a}^2({\dot{\alpha }}^2+2\dot{\alpha }\dot{\beta }\cos (\beta )+{\dot{\beta }}^2)$. The Lagrangian of the system is the difference between the kinetic and potential energies: $\mathcal{L}=T-V$.

Linearize the equations

To linearize the equations, we assume that $\alpha ,\beta \ll 1$, so we can approximate the potentials and velocities in the Lagrangian equation as follows: $V\approx -mg(2a-\frac{1}{2}a{\alpha }^2-a\alpha \beta )$ $T\approx \frac{1}{2}m{a}^2({\dot{\alpha }}^2+{\dot{\beta }}^2)$

Apply the small-angle approximation

The linearized Lagrangian is now $\mathcal{L}=\frac{1}{2}m{a}^2({\dot{\alpha }}^2+{\dot{\beta }}^2)+mg(a{\alpha }^2+2a\alpha \beta )$. Euler-Lagrange equations give us: $\ddot{\alpha }=-g(\alpha +\beta )$ $\ddot{\beta }=-g\alpha$

Find the normal frequencies of oscillation

We now have a system of linearized differential equations. To find the normal frequencies, we assume solutions of the form $\alpha =A{e}^{i\omega t}$ and $\beta =B{e}^{i\omega t}$. Substituting these solutions back into the differential equations, we get: $-A{\omega }^2=-g(A+B)$ $-B{\omega }^2=-gA$ Solving for the eigenvalues ${\omega }^2$ using matrix algebra, we find that the normal frequencies are: ${\omega }_{1,2}=\sqrt{(2\pm \sqrt{2})g/a}$

Describe the corresponding normal modes

The normal modes correspond to the eigenvectors of the matrix equation: For ${\omega }_1$: - The first normal mode has both bobs oscillating in phase, which means they move in the same direction simultaneously. - The amplitude of oscillation relative to each other is equal. For ${\omega }_2$: - The second normal mode has the bobs oscillating out of phase, which means they move in the opposite direction. - The second bob oscillates with a higher amplitude than the first bob. Now, to describe the normal modes when the motion is not confined to a vertical plane, we can make qualitative observations without doing any further calculations: When the motion is not confined to a vertical plane, the normal modes will still consist of in-phase and out-of-phase oscillations, but the bobs will be free to oscillate in different planes. The in-phase mode will have both bobs moving in a circular motion in the same direction, while the out-of-phase mode will have the bobs moving in opposite directions.

Key concepts

Normal Frequencies

When studying the dynamics of systems like the double pendulum, the concept of normal frequencies is crucial. These are the natural frequencies at which the system tends to oscillate in the absence of external forces. Think of them as the pendulum's unique 'heartbeat' when it swings back and forth.

In the case of a double pendulum under the influence of gravity, finding these frequencies involves applying physics principles and a bit of mathematical legwork. By assuming that both masses are identical and the system oscillates near its equilibrium point, we can apply linearized equations of motion to determine the natural frequencies of the system. For the double pendulum, this process reveals two different frequencies, which represent two distinct ways the pendulum can oscillate, known as the normal modes.

To visualize the normal modes, imagine one where both bobs swing together in harmony (in-phase mode) and another where the bobs swing in opposition (out-of-phase mode), with the second bob moving more vigorously. It's like two dancers performing together — one mode shows them moving in unison, while the other mode has them executing complementary moves.

Lagrangian Mechanics

Delving into the world of Lagrangian mechanics might seem like a journey through a dark forest for the uninitiated, but it's a powerful tool for understanding systems like our double pendulum. Lagrangian mechanics is essentially an elegant reformulation of classical mechanics that focuses on energy rather than forces. The central player here is the Lagrangian, denoted by $\mathcal{L}$, which is the difference between the kinetic energy (T) and potential energy (V) of a system.

This approach simplifies the motion equations for complex systems. Through the Euler-Lagrange equations – magical equations that apply the principle of least action – we derive equations of motion that describe the dynamics of the system. In the case of the double pendulum, using the Lagrangian allows simplifying the problem by considering energies rather than tracking every push and pull acting on the bobs. By doing this, we can elegantly navigate the dynamics of the system and predict its behavior.

Small-Angle Approximation

In mechanics, sometimes we must turn to simplifying assumptions to crack tough nuts like the double pendulum's equations. Enter the small-angle approximation, a simplification often used when dealing with pendulum-related problems. What's the gist of it? Well, when the angles involved are small, the sine and tangent functions of the angle are approximately equal to the angle itself (when measured in radians).

Here's why it's important: small-angle approximations transform the sinusoidal terms in the equations of motion into polynomial ones, which are much simpler to handle. This nifty trick strips down the complexity of the original equations drastically. Back to our double pendulum, by assuming that the angles ${\theta }_1$ and ${\theta }_2$ are small, we manage to linearize the problem, leading us to results that would otherwise require complex, nonlinear differential equations. Despite being an approximation, it's surprisingly effective for near-equilibrium motions, providing accurate results without the mathematical headache.