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Chapter 13: Problem 61

Show that the potential energy of a simple pendulum is proportional to the square of the angular displacement in the small-amplitude limit.

Short Answer

Expert verified
We showed that the potential energy of a simple pendulum with a small amplitude displacement is proportional to the square of the angular displacement \(\theta^2\).

Step by step solution

01

Define the Potential Energy (PE)

The potential energy of a pendulum upswing from the equilibrium position to some maximum angle \(\theta\) is given by PE = m * g * h. Here, m is the mass of the pendulum bob, g is the acceleration due to gravity, and h is the vertical height risen by the pendulum.
02

Expressing the vertical height (h) in terms of angular displacement \(\theta\)

The height h can be expressed as h = L - Lcos\(\theta\), where L is the length of the pendulum. When the pendulum is at maximum displacement, it forms a right triangle with the pendulum鈥檚 arm. The hypotenuse is L, and cos\(\theta\) is (L - h) / L, thus h = L - Lcos\(\theta\).
03

Substituting h in the Potential Energy (PE) equation

Substitute h in the PE equation, we have PE = m * g * (L - Lcos\(\theta\)). Simplifying further gives PE = m * g * L - m * g * Lcos\(\theta\). The first term m * g * L is the constant potential energy when the pendulum is undisturbed. Therefore, we focus on the term with cos\(\theta\).
04

Using the small angle approximation and finalizing the equation

For small angles, cos\(\theta\) \approx 1 - \(\theta^2\) / 2. Substituting back into the equation, the potential energy is PE = m * g * L - m * g * L * (1 - \(\theta^2\) / 2) = m * g * L * \(\theta^2\) / 2. The potential energy is therefore proportional to \(\theta^2\), which completes the demonstration.

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