Problem 25 - A First Course in Differential Equations with Modeling Applications

Question

Motion in a Force Field \(\quad\) A mathematical model for the position \(x(t)\) of a body moving rectilinearly on the \(x\) -axis in an inverse-square force field is given by \\[\frac{d^{2} x}{d t^{2}}=-\frac{k^{2}}{x^{2}}\\] Suppose that at \(t=0\) the body starts from rest from the position \(x=x_{0}, x_{0}>0 .\) Show that the velocity of the body at time \(t\) is given by \(v^{2}=2 k^{2}\left(1 / x-1 / x_{0}\right) .\) Use the last expression and a CAS to carry out the integration to express time \(t\) in terms of \(x\)

Step-by-Step Solution

Answer

Velocity is \( v^2 = 2k^2 \left( \frac{1}{x} - \frac{1}{x_0} \right) \). Integrate to express time in terms of \( x \).

Step 1: Understand the Problem

We have a second-order differential equation \( \frac{d^{2} x}{d t^{2}} = -\frac{k^{2}}{x^{2}} \) describing the motion in an inverse-square force field. Our objective is to prove the expression for velocity \( v \) and use it to find an expression for time \( t \) in terms of position \( x \).

Step 2: Setup the Differential Equation

Since the force is given by \( F = m \frac{d^{2} x}{d t^{2}} \), and force is \( F = - \frac{k^{2}}{x^{2}} \), we equate the accelerations: \( \frac{d^{2} x}{d t^{2}} = - \frac{k^{2}}{x^{2}} \). Start by integrating this with respect to \( x \) to find the velocity expression.

Step 3: Integrate to Find Velocity

To find velocity, integrate the acceleration equation \( \frac{d^{2} x}{d t^{2}} = - \frac{k^{2}}{x^{2}} \). Use the chain rule: \( \frac{d^{2} x}{d t^{2}} = \frac{d}{dt}(\frac{dx}{dt}) = \frac{d}{dx}(\frac{dx}{dt}) \cdot \frac{dx}{dt} = \frac{d v}{dx} \cdot v \). This gives \( v \frac{dv}{dx} = - \frac{k^{2}}{x^{2}} \).

Step 4: Solve for the Velocity Squared

Integrate \( v \, dv = - \frac{k^{2}}{x^{2}} dx \). This leads to \( \int v \, dv = - \int \frac{k^{2}}{x^{2}} dx \). The left side integrates to \( \frac{v^2}{2} \), and the right to \( k^{2} \frac{1}{x} + C \), resulting in \( \frac{v^2}{2} = k^{2} \left( \frac{1}{x} \right) + C \).

Step 5: Use Initial Conditions

Use initial conditions: At \( t = 0, x = x_0 \) and the body is at rest, so \( v = 0 \). Substitute \( x_0 \) and \( v = 0 \) into \( \frac{v^2}{2} = k^{2} \left( \frac{1}{x} \right) + C \) to find \( C \). This gives \( C = -k^2 \frac{1}{x_0} \).

Step 6: Express Velocity

Substitute \( C \) back: \( \frac{v^2}{2} = k^{2} \left( \frac{1}{x} - \frac{1}{x_0} \right) \), leading to \( v^2 = 2k^2 \left( \frac{1}{x} - \frac{1}{x_0} \right) \), which is the required velocity expression.

Step 7: Relate Velocity to Time and Position

With the expression for \( v \), the velocity is \( v = \frac{dx}{dt} \). So, \( \sqrt{2k^2 \left( \frac{1}{x} - \frac{1}{x_0} \right)} = \frac{dx}{dt} \). Rearrange to separate variables: \( dt = \frac{dx}{\sqrt{2k^2 \left( \frac{1}{x} - \frac{1}{x_0} \right)}} \).

Step 8: Integrate to Find Time

Integrate both sides: \( \int dt = \int \frac{dx}{\sqrt{2k^2 \left( \frac{1}{x} - \frac{1}{x_0} \right)}} \). This integration can be solved using a Computer Algebra System (CAS) to find the time \( t \) in terms of position \( x \).

Key Concepts

Inverse-Square Force Field

In an inverse-square force field, the force acting on an object is inversely proportional to the square of the distance from a source point. This type of force field is a classic model in physics, often encountered in gravitational and electrostatic fields.

The force can be expressed mathematically as:

• \( F = -\frac{k^2}{x^2} \)

where \( F \) is the force exerted, \( k \) is a constant, and \( x \) is the distance from the source.

This model describes how the influence of the force decreases rapidly as the distance \( x \) increases. The inverse-square law is crucial to understanding how objects move under such forces, leading to various real-world applications like planetary motion.

Second-Order Differential Equation

A second-order differential equation involves the second derivative of a variable. In this exercise, the position \( x(t) \) is described by the equation:

• \( \frac{d^2 x}{dt^2} = -\frac{k^2}{x^2} \)

This equation is key to modeling the dynamics of an object in an inverse-square force field.

The second derivative, \( \frac{d^2 x}{dt^2} \), represents acceleration. This is essential when dealing with forces, as forces directly impact acceleration according to Newton's second law. To solve such equations, one typically integrates twice, as knowing acceleration allows us to find velocity and then position.

Integration Technique

Integrating a differential equation is crucial in finding the expressions for velocity and position from acceleration. Let's look at how integration is applied in this problem:

• The equation \( v \frac{dv}{dx} = -\frac{k^2}{x^2} \) is derived using the chain rule. It allows us to move from acceleration to knowing velocity behavior.
• The next step is integrating both sides: \( \int v \, dv = -\int \frac{k^2}{x^2} \, dx \).

Upon integration, we get:

• \( \frac{v^2}{2} = k^2 \left( \frac{1}{x} \right) + C \)

By correctly using the initial conditions, the constant \( C \) is determined, leading to a useful expression for velocity. Integration thus links dynamics to initial states, playing a pivotal role in solution methods for differential equations.

Initial Conditions

Initial conditions specify the state of the system at the start of examination. For this exercise:

• At time \( t=0 \), the position is \( x = x_0 \) and the velocity \( v = 0 \).

These conditions are crucial for finding the integration constant \( C \).

When substituting into the velocity equation \( \frac{v^2}{2} = k^2 \left( \frac{1}{x} \right) + C \), the initial conditions allow us to find that:

• \( C = -k^2 \frac{1}{x_0} \)

Initial conditions thus anchor the solution to a real-world scenario, enabling us to connect abstract mathematical solutions with tangible physical systems.