Question

Consider the motion of a particle of mass in falling vertically under the earth's gravitational field, and suppose the downward motion is opposed by a frictional force \(p(v)\) dependent on the velocity \(v(t)\) of the particle. Then the velocity satisfies the equation $$ m v^{\prime}(t)=-m g+p(v), \quad t \geq 0, \quad v(0) \text { given } $$ Let \(m=1 \mathrm{~kg}, g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\), and \(v(0)=0\). Solve the differential equation for \(0 \leq t \leq 20\) and for the following choices of \(p(v)\) : (a) \(p(v)=-0.1 v\), which is positive for a falling body. (b) \(\quad p(v)=0.1 v^{2}\). Find answers to at least three digits of accuracy. Graph the functions \(v(t) .\) Compare the solutions.

Short answer

For (a), the solution is \( v(t) = -98 + 98e^{-0.1t} \). For (b), the solution is \( v(t) = \sqrt{980e^{\frac{t}{10}} - 980} \). Plot and compare both.

Step-by-step solution

Understanding the Differential Equation

Given the differential equation for the particle's velocity: \[ m v^{\prime}(t) = -m g + p(v) \] with initial conditions: \[ m = 1 \text{ kg}, \ g = 9.8 \text{ m/sec}^2, \ v(0) = 0 \] It can be simplified as: \[ v^{\prime}(t) = -9.8 + p(v) \]

Case (a) Solve for p(v) = -0.1v

Substitute \( p(v) = -0.1v \) into the simplified differential equation: \[ v^{\prime}(t) = -9.8 - 0.1v \] This is a first-order linear differential equation. To solve it, use the integrating factor method. The standard form is: \[ v^{\prime} + 0.1v = -9.8 \] The integrating factor \( \mu(t) \) is: \[ \mu(t) = e^{\int 0.1 dt} = e^{0.1t} \] Multiplying through by the integrating factor gives: \[ e^{0.1t} v^{\prime} + 0.1 e^{0.1t} v = -9.8 e^{0.1t} \] The left side is the derivative of \( e^{0.1t} v \), so: \[ \frac{d}{dt} (e^{0.1t} v) = -9.8 e^{0.1t} \] Integrate both sides: \[ e^{0.1t} v = -9.8 \int e^{0.1t} dt = -9.8 \frac{e^{0.1t}}{0.1} + C = -98 e^{0.1t} + C \] Solve for \( v \): \[ v(t) = -98 + Ce^{-0.1t} \] Using the initial condition \( v(0) = 0 \), solve for \( C \): \[ 0 = -98 + C \implies C = 98 \] Thus, the solution is: \[ v(t) = -98 + 98e^{-0.1t} \]

Case (b) Solve for p(v) = 0.1v^2

Substitute \( p(v) = 0.1v^2 \) into the simplified differential equation: \[ v^{\prime}(t) = -9.8 + 0.1v^2 \] This is a nonlinear differential equation and can be solved using separation of variables. Write it in separable form: \[ \frac{dv}{-9.8 + 0.1v^2} = dt \] Integrate both sides. Let \( v^2 = \frac{u}{0.1} = 10u \), so \( dv = \frac{du}{10v} \): \[ \int \frac{10}{-9.8 + u} du = \int dt \] Solving the integrals, we get: \[ 10 \int \frac{1}{-9.8 + u} du = t + C \] The left integral is: \[ 10 \ln| -9.8 + u | = t + C \] Reverting back to \( v \): \[ 10 \ln| -9.8 + 0.1v^2 | = t + C \] Exponentiating both sides: \[ -9.8 + 0.1v^2 = e^{\frac{t}{10} + C} = Ae^{\frac{t}{10}} \] Solving for \( v \): \[ 0.1v^2 = -9.8 + Ae^{\frac{t}{10}} \implies v(t) = \sqrt{10(Ae^{\frac{t}{10}} - 98)} \] Using the initial condition \( v(0) = 0 \) to find \( A \): \[ 0 = \sqrt{10(A - 98)} \implies A = 98 \] Thus, the solution is: \[ v(t) = \sqrt{10(98e^{\frac{t}{10}} - 98)} = \sqrt{980e^{\frac{t}{10}} - 980} \]

Graph the Functions and Compare Solutions

Plot the functions \( v(t) = -98 + 98e^{-0.1t} \) and \( v(t) = \sqrt{980e^{\frac{t}{10}} - 980} \) for \( 0 \leq t \leq 20 \). Comparing both solutions, observe the decay behavior in the first case (a) and the nonlinear growth in the second case (b).

Key concepts

Gravitational Motion

Gravitational motion describes the movement of objects under the influence of gravity. For a particle falling towards Earth, gravity is the primary force acting on it. The gravitational force on a particle of mass is given by Newton's law of universal gravitation, simplified as: \[ F = mg \] where:

• \(F\) is the force due to gravity,
• \(m\) is the mass of the particle, and
• \(g\) is the acceleration due to gravity (approximately \(9.8 \, \text{m/s}^{2}\) near the Earth's surface).

In this exercise, the motion is also influenced by friction, which is a resistive force dependent on the particle's velocity. The falling particle's downward velocity thus satisfies a differential equation that accounts for both gravitational pull and frictional resistance.

First-Order Linear Differential Equations

A first-order linear differential equation has the general form: \[ \frac{dv}{dt} + P(t)v = Q(t) \] where \(P(t)\) and \(Q(t)\) are given functions of \(t\). In this particular problem, when \(p(v) = -0.1v\), the differential equation becomes: \[ \frac{dv}{dt} = -9.8 - 0.1v \] Rewriting in the standard form: \[ \frac{dv}{dt} + 0.1v = -9.8 \] This format is suitable for solving using methods such as the integrating factor method, which helps in finding the general solution to these kinds of equations efficiently.

Nonlinear Differential Equations

Nonlinear differential equations contain terms that are non-linear with respect to the unknown function or its derivatives. In the given exercise, when \(p(v) = 0.1v^2\), the equation becomes: \[ \frac{dv}{dt} = -9.8 + 0.1v^2 \] This is a nonlinear differential equation. Solving nonlinear differential equations often requires different methods compared to linear ones, like separation of variables, which helps isolate the terms involving the unknown function. Nonlinear equations can exhibit more complex behavior and solutions compared to their linear counterparts.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations. The idea is to multiply the differential equation by an integrating factor, which simplifies the equation into an easily integrable form. For the equation: \[ \frac{dv}{dt} + 0.1 v = -9.8 \] The integrating factor \( \mu(t) \) is calculated as: \[ \mu(t) = e^{\int 0.1 dt} = e^{0.1t} \] Multiplying both sides of the equation by \( \mu(t) \) transforms it into: \[ \frac{d}{dt} (e^{0.1t} v) = -9.8 e^{0.1t} \] Now the equation can be integrated easily, allowing us to solve for \(v(t)\).

Separation of Variables

Separation of variables is a method for solving differential equations by separating the variables (unknown function and its derivative) on different sides of the equation. For the nonlinear equation: \[ \frac{dv}{dt} = -9.8 + 0.1v^2 \] We rewrite it in a separable form: \[ \frac{dv}{-9.8 + 0.1v^2} = dt \] Then, both sides of the equation can be integrated individually. This technique helps in tackling the complexity of nonlinear differential equations and finding their solutions by breaking down the problem into simpler integrable parts. Keeping track of integrating limits and initial conditions is crucial for finding the appropriate constant and thus the exact solution.