Question
What is the differential equation for a pendulum's motion in a resistive medium?
Answers
GPT-4o mini
The motion of a pendulum experiencing resistive forces can be described by a second-order differential equation. In a simple pendulum, the forces acting on the pendulum bob include gravitational force and a resistive (damping) force due to the medium through which it moves (like air resistance or friction).
Let’s denote:
- \( \theta \): the angular displacement from the vertical (in radians),
- \( m \): the mass of the pendulum bob,
- \( L \): the length of the pendulum,
- \( g \): the acceleration due to gravity,
- \( b \): the damping coefficient (a constant representing the resistive force),
- \( I \): the moment of inertia of the pendulum bob (for a point mass \( m \) at length \( L \), \( I = mL^2 \)).
The equation of motion for a simple pendulum (without damping) can be written as:
\[
I \frac{d^2\theta}{dt^2} + m g L \sin(\theta) = 0.
\]
However, when we include damping, we need to add a term representing the resistive force. The resistive force can be assumed to be proportional to the angular velocity \( \frac{d\theta}{dt} \), and can be expressed as \( -b \frac{d\theta}{dt} \).
Thus, the modified equation of motion becomes:
\[
I \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + m g L \sin(\theta) = 0.
\]
Substituting \( I = m L^2 \), we get:
\[
m L^2 \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + m g L \sin(\theta) = 0.
\]
Dividing through by \( m L^2 \) (assuming \( m \neq 0 \) and \( L \neq 0 \)):
\[
\frac{d^2\theta}{dt^2} + \frac{b}{m L^2} \frac{d\theta}{dt} + \frac{g}{L} \sin(\theta) = 0.
\]
Define \( \gamma = \frac{b}{m L^2} \) and \( \omega_0^2 = \frac{g}{L} \) (the natural frequency squared). The final form of the differential equation for a pendulum's motion in a resistive medium is then:
\[
\frac{d^2\theta}{dt^2} + \gamma \frac{d\theta}{dt} + \omega_0^2 \sin(\theta) = 0.
\]
This equation describes a damped pendulum, where the damping term reduces the amplitude of the oscillations over time due to the resistive medium.
Let’s denote:
- \( \theta \): the angular displacement from the vertical (in radians),
- \( m \): the mass of the pendulum bob,
- \( L \): the length of the pendulum,
- \( g \): the acceleration due to gravity,
- \( b \): the damping coefficient (a constant representing the resistive force),
- \( I \): the moment of inertia of the pendulum bob (for a point mass \( m \) at length \( L \), \( I = mL^2 \)).
The equation of motion for a simple pendulum (without damping) can be written as:
\[
I \frac{d^2\theta}{dt^2} + m g L \sin(\theta) = 0.
\]
However, when we include damping, we need to add a term representing the resistive force. The resistive force can be assumed to be proportional to the angular velocity \( \frac{d\theta}{dt} \), and can be expressed as \( -b \frac{d\theta}{dt} \).
Thus, the modified equation of motion becomes:
\[
I \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + m g L \sin(\theta) = 0.
\]
Substituting \( I = m L^2 \), we get:
\[
m L^2 \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + m g L \sin(\theta) = 0.
\]
Dividing through by \( m L^2 \) (assuming \( m \neq 0 \) and \( L \neq 0 \)):
\[
\frac{d^2\theta}{dt^2} + \frac{b}{m L^2} \frac{d\theta}{dt} + \frac{g}{L} \sin(\theta) = 0.
\]
Define \( \gamma = \frac{b}{m L^2} \) and \( \omega_0^2 = \frac{g}{L} \) (the natural frequency squared). The final form of the differential equation for a pendulum's motion in a resistive medium is then:
\[
\frac{d^2\theta}{dt^2} + \gamma \frac{d\theta}{dt} + \omega_0^2 \sin(\theta) = 0.
\]
This equation describes a damped pendulum, where the damping term reduces the amplitude of the oscillations over time due to the resistive medium.