Question

Consider a simple pendulum undergoing simple harmonic motion with a time period \(T\), and a fixed amplitude \(\theta_0\) of angular oscillation. Its angular momentum about the point of suspension exhibits an oscillatory behavior with an amplitude \(A\). Which of the following relations between \(A\) and \(T\) is correct?

• A. \(A \propto T^3\)
• B. \(A \propto T^2\)
• C. \(A \propto T\)
• D. \(A \propto T^4\)

Hint:

Express all variables in terms of the primary variable \(T\).
Length \(L \propto T^2\), which means moment of inertia \(I \propto L^2 \propto T^4\).
Since angular momentum amplitude is \(I \omega\) and \(\omega \propto T^{-1}\), we directly get \(T^4 \cdot T^{-1} = T^3\).

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the angular momentum of a pendulum and its time period. A simple pendulum undergoes simple harmonic motion characterized by its angular oscillation and angular momentum.

1. The time period \( T \) of a simple pendulum is given by the formula: \(T = 2\pi\sqrt{\frac{l}{g}}\) where \( l \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
2. The angular momentum \( L \) about the point of suspension is related to the mass \( m \), the length \( l \), and the angular velocity \( \omega \). It can be expressed as: \(L = I \omega = ml^2 \omega\) where \( I = ml^2 \) is the moment of inertia.
3. For small angular displacements, the angular velocity \( \omega \) can be expressed in terms of the maximum angular displacement \( \theta_0 \): \(\omega = \frac{d\theta}{dt} \approx -\omega_0 \sin(\omega_0 t) = -\frac{2\pi}{T} \sin\left(\frac{2\pi t}{T}\right)\) where \( \omega_0 = \frac{2\pi}{T} \) is the angular frequency.
4. The amplitude of angular momentum \( A \) is proportional to \( ml^2 \frac{2\pi}{T} \theta_0 \), so we have: \(A \propto ml^2 \theta_0 \frac{2\pi}{T}\) Replace \( l \) using the time period formula: \(l = \frac{T^2 g}{4\pi^2}\)
5. Substitute \( l \) in the amplitude expression: \(A \propto m \left(\frac{T^2 g}{4\pi^2}\right)^2 \theta_0 \frac{2\pi}{T} = \frac{m g^2 \theta_0}{8\pi^3} T^3\)
6. Therefore, the amplitude \( A \) of the angular momentum is proportional to the cube of the time period \( T \): \(A \propto T^3\)

Thus, the correct relation is that the amplitude of angular momentum is proportional to the cube of the time period. Therefore, the correct answer is:

\(A \propto T^3\)