Question

Stable states. For the energy function \(V(\theta)=\cos \theta\) for \(0 \leq \theta \leq 2 \pi\), find the values \(\theta=\theta_{s}\) that identify stable equilibria, and the values \(\theta=\theta_{u}\) that identify unstable equilibria.

Short answer

Stable equilibrium: \( \theta_{s} = \pi \). Unstable equilibria: \( \theta_{u} = 0, 2\pi \).

Step-by-step solution

Find the First Derivative

To find the stable and unstable equilibria, first calculate the first derivative of the energy function. The energy function is given as \[ V(\theta) = \cos \theta \]. The first derivative with respect to \(\theta\) is \[ V'(\theta) = -\sin \theta \].

Set the First Derivative to Zero

Determine the critical points by setting the first derivative equal to zero: \[ -\sin \theta = 0 \]. This gives us \[ \sin \theta = 0 \]. The solutions to \(\sin \theta = 0\) in the interval \(0 \leq \theta \leq 2 \pi\) are: \[ \theta = 0, \ \theta = \pi \ and \ \theta = 2\pi \].

Find the Second Derivative

To determine if these points are stable or unstable, compute the second derivative of the energy function. The first derivative is \[ V'(\theta) = -\sin \theta \]. The second derivative with respect to \(\theta\) is \[ V''(\theta) = -\cos \theta \].

Evaluate the Second Derivative at Critical Points

Evaluate the second derivative at each critical point to determine the concavity and classify the equilibrium: 1. At \( \theta = 0 \):\[ V''(0) = -\cos(0) = -1 \] (negative, so it's a maximum, unstable equilibrium)2. At \( \theta = \pi \):\[ V''(\pi) = -\cos(\pi) = 1 \] (positive, so it's a minimum, stable equilibrium)3. At \( \theta = 2\pi \):\[ V''(2\pi) = -\cos(2\pi) = -1 \] (negative, so it's a maximum, unstable equilibrium).

Identify Stable and Unstable Equilibria

From the second derivative test, identify the stable and unstable equilibria: - Stable equilibrium: \( \theta_{s} = \pi \)- Unstable equilibria: \( \theta_{u} = 0, 2\pi \).

Key concepts

Critical Points

Critical points are essential for finding where a function's slope is zero. These points are found by setting the first derivative of the function equal to zero. For our energy function, we have:
\[ V(\theta) = \cos(\theta) \],
The first derivative is:
\[ V'(\theta) = -\sin(\theta) \].
Setting this equal to zero gives us our critical points:
\[ -\sin(\theta) = 0 \].
So, \[ \theta = 0, \pi, 2\pi \].
Critical points help us identify spots to test for stability. They can be transitions between increasing and decreasing values, indicating possible peaks and troughs of the function.

First Derivative Test

The first derivative test involves examining the first derivative (slope) of a function. If the slope (first derivative) is zero at a point, that point is critical. The function changes direction at these critical points.
For our function:
\[ V(\theta) = \cos(\theta) \],
The first derivative is:
\[ V'(\theta) = -\sin(\theta) \].
By setting the first derivative to zero:
\[ -\sin(\theta) = 0 \],
we find the critical points:
\[ \theta = 0, \pi, 2\pi \].
The first derivative changes signs at these points, indicating either a peak or a trough in the energy function.

Second Derivative Test

The second derivative test helps determine the concavity at critical points and whether these points are maxima, minima, or inflection points. For our energy function:
\[ V'(\theta) = -\sin(\theta) \],
The second derivative is:
\[ V''(\theta) = -\cos(\theta) \].
Evaluate the second derivative at each critical point to classify:

• At \theta = 0: \[ V''(0) = -\cos(0) = -1 \] (concave down, unstable)
• At \theta = \pi: \[ V''(\pi) = -\cos(\pi) = 1 \] (concave up, stable)
• At \theta = 2\pi: \[ V''(2\pi) = -\cos(2\pi) = -1 \] (concave down, unstable)

Once we know the concavity, we can tell which points are stable or unstable equilibrium.

Stable Equilibrium

A stable equilibrium occurs at a point where the second derivative is positive. This means the function is concave up, and any small disturbance will lead the system to return to this point. From our function:
\[ V(\theta) = \cos(\theta) \],
we have:

• At \theta = \pi, \ V''(\theta) = 1 \ is positive, indicating stability.

This makes \theta = \pi a stable equilibrium. The system tends to settle here, as it represents a local minimum in the energy function. Disturbing it slightly will still lead it to come back to this point.

Unstable Equilibrium

Unstable equilibrium occurs where the second derivative is negative, indicating concave down. Small disturbances lead away from this point. For our function:
\[ V(\theta) = \cos(\theta) \],
we have:

• At \theta = 0, \ V''(0) = -1 \ is negative, showing instability.
• At \theta = 2\pi, \ V''(2\pi) = -1 \ is also negative, showing instability.

These points are unstable, representing local energy maxima. Any disturbance makes the system move away from these points, pushing it towards a more stable state.