Problem 25 - Advanced Engineering Mathematics

Question

A force of 2 pounds stretches a spring 1 foot. A mass weighing \(3.2\) pounds is attached to the spring, and the system is then immersed in a medium that offers a damping force numerically equal to \(0.4\) times the instantaneous velocity. (a) Find the equation of motion if the mass is initially released from rest from a point 1 foot above the equilibrium position. (b) Express the equation of motion in the form given in (23). (c) Find the first time at which the mass passes through the equilibrium position heading upward.

Step-by-Step Solution

Answer

The equation of motion is \( x(t) = e^{-2t}(-\cos 4t - \frac{1}{2} \sin 4t) \). The mass first passes upward through equilibrium at \( t \approx 0.1121 \) seconds.

Step 1: Determine Spring Constant

According to Hooke's Law, the force exerted by a spring is proportional to the displacement. Thus, we have \( F = kx \) where \( F = 2 \) pounds and \( x = 1 \) foot. Therefore, \( k = \frac{F}{x} = \frac{2}{1} = 2 \) pounds per foot.

Step 2: Setup Differential Equation

The mass is attached to a spring and experiences both a spring force and a damping force. The equation of motion is given by:\[m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0\]Given \( m = \frac{3.2}{32} = 0.1 \) slugs, \( c = 0.4 \), and \( k = 2 \), this becomes:\[\0.1\frac{d^2x}{dt^2} + 0.4\frac{dx}{dt} + 2x = 0.\]

Step 3: Simplify the Differential Equation

Divide the entire equation by the mass \( 0.1 \) to normalize it:\[\\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 20x = 0.\]

Step 4: Solve for General Solution

The characteristic equation is \( r^2 + 4r + 20 = 0 \). Solving using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we get:\[\r = \frac{-4 \pm \sqrt{16 - 80}}{2} = -2 \pm 4i.\]This implies a general solution of the form \( x(t) = e^{-2t}(C_1 \cos 4t + C_2 \sin 4t) \).

Step 5: Apply Initial Conditions

Given initial conditions: released from rest \( x(0) = -1 \) (1 foot above equilibrium is negative) and \( \frac{dx}{dt}(0) = 0 \), solve the system:1. \( x(0) = -1 = C_1 \)2. \( \frac{dx}{dt}(0) = -2C_1 + 4C_2 = 0 \rightarrow -2(-1) + 4C_2 = 0 \rightarrow 2 + 4C_2 = 0 \therefore C_2 = -\frac{1}{2} \)Thus, \( x(t) = e^{-2t}(-\cos 4t - \frac{1}{2} \sin 4t) \).

Step 6: Express Equation in Given Form

Per part (b) requirements, we express the equation as:\( x(t) = e^{-2t}(-\cos 4t - \frac{1}{2} \sin 4t) \), which is consistent with the general solution structure.

Step 7: Find First Zero of x(t)

To find when the mass first passes through equilibrium (\( x(t) = 0 \) and \( \frac{dx}{dt} > 0 \)), solve:\[\-e^{-2t}(\cos 4t + \frac{1}{2} \sin 4t) = 0\rightarrow \cos 4t + \frac{1}{2} \sin 4t = 0.\]By \( \tan 4t = -2 \), solve for \( t \). The smallest positive solution is when:\ \( 4t = \tan^{-1}(-2) + n\pi \) for the first \( n = 0 \):\\( t = \frac{\tan^{-1}(-2)}{4} \approx 0.1121 \). Hence, the first time it crosses equilibrium going upward is at approximately \( t = 0.1121 \) seconds.

Key Concepts

Spring Constant

When working with springs, understanding the spring constant, denoted as \( k \), is key. The spring constant is a measure of a spring's stiffness and dictates how much force is needed to stretch or compress the spring by a unit distance. In mathematical terms, this is expressed by Hooke's Law:

• \( F = kx \)

where \( F \) is the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring's natural length.
In the exercise provided, we're told a force of 2 pounds stretches the spring 1 foot. This allows us to calculate the spring constant using:

• \( k = \frac{F}{x} = \frac{2}{1} = 2 \) pounds per foot

This means every foot the spring is stretched requires a force of 2 pounds.

Differential Equations

Differential equations are equations that involve derivatives. They are crucial for modeling how physical systems change over time. When a system experiences both elastic and damping forces, the motion can be described using a second-order linear differential equation:

• \( m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \)

Here, \( m \) represents the mass, \( c \) is the damping coefficient, and \( k \) is the spring constant. The equation works by balancing these forces to describe the acceleration and velocity of a mass on a spring over time.
In our exercise, once the mass and damping forces are incorporated, and the equation is further simplified by dividing by mass, it becomes:

• \( \frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 20x = 0 \)

Initial Conditions

Initial conditions specify the starting state of a system to solve a differential equation accurately. They provide the necessary context to tailor the general solution to a specific situation.
For instance, initial conditions in this problem indicate that the mass is released from rest at a position 1 foot above equilibrium. This translates to:

• \( x(0) = -1 \)
• \( \frac{dx}{dt}(0) = 0 \)

The initial position being negative signifies 1 foot above equilibrium (considering downward as positive).
Applying these conditions to the general solution \(x(t) = e^{-2t}(C_1 \cos 4t + C_2 \sin 4t)\), we determine the specific constants \(C_1\) and \(C_2\) for this situation.

Characteristic Equation

The characteristic equation arises from the process of solving differential equations, specifically those describing harmonic motion. It is obtained by assuming a trial solution of the exponential form and substituting it into the differential equation.
For our simplified equation:

• \( \frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 20x = 0 \)

The characteristic equation is:

• \( r^2 + 4r + 20 = 0 \)

Solving this quadratic equation using the quadratic formula yields:

• \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = -2 \pm 4i \)

The solution \(r = -2 \pm 4i\) suggests the system is underdamped, leading to oscillations with an exponential decay factor. This information is crucial for predicting the system's behavior over time.