Problem 93 - Calculus Volume 3

Question

An 800 -lb weight ( 25 slugs) is attached to a vertical spring with a spring constant of \(226 \mathrm{lb} / \mathrm{ft}\). The system is immersed in a medium that imparts a damping force equal to 10 times the instantaneous velocity of the mass. a. Find the equation of motion if it is released from a position \(20 \mathrm{ft}\) below its equilibrium position with a downward velocity of \(41 \mathrm{ft} / \mathrm{sec}\). b. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

Step-by-Step Solution

Answer

Equation of motion: \( x(t) = e^{-0.2t}(-20 \cos(3t) - 13 \sin(3t)) \). Motion is underdamped.

Step 1: Convert Units

First, convert the weight of the object to mass in slugs. The given weight is 800 lb. Since weight is the product of mass and gravity, we find the mass using the relation \( m = \frac{W}{g} \), where \( W = 800 \) lb and \( g = 32 \frac{ft}{s^2} \). Therefore, the mass \( m \) is 25 slugs.

Step 2: Identify Given Data

We have the mass \( m = 25 \) slugs, spring constant \( k = 226 \) lb/ft, and damping coefficient \( c = 10 \) lb·s/ft (since the damping force is 10 times the velocity). Initially, the object is displaced 20 ft below equilibrium, \( x(0) = -20 \) ft, and it has a downward velocity of 41 ft/s, \( \frac{dx}{dt}(0) = -41 \).

Step 3: Set Up Differential Equation

The differential equation for damped harmonic motion is \( m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \). Substituting the values, we obtain: \[ 25\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 226x = 0 \].

Step 4: Solve Characteristic Equation

The characteristic equation corresponding to the differential equation is \( 25r^2 + 10r + 226 = 0 \). Using the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we substitute \( a = 25 \), \( b = 10 \), \( c = 226 \).

Step 5: Calculate Roots

Calculate the discriminant: \( b^2 - 4ac = 10^2 - 4(25)(226) = 100 - 22600 = -22500 \). Since the discriminant is negative, the roots are complex: \( r = \frac{-10 \pm i\sqrt{22500}}{50} = \frac{-10 \pm 150i}{50} \). Thus, the roots are \( r = -0.2 \pm 3i \).

Step 6: Formulate General Solution

The general solution to the damped harmonic motion is \( x(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \), where \( \alpha = -0.2 \) and \( \beta = 3 \). So the solution becomes \( x(t) = e^{-0.2t}(C_1 \cos(3t) + C_2 \sin(3t)) \).

Step 7: Apply Initial Conditions

Use the initial conditions \( x(0) = -20 \) and \( \frac{dx}{dt}(0) = -41 \) to solve for \( C_1 \) and \( C_2 \). Substituting \( t = 0 \) in the general solution: \( -20 = C_1 \). Next, differentiate \( x(t) \) and substitute into the derivative with \( t = 0 \) to find \( C_2 \).

Step 8: Determine C_1 and C_2

Differentiating \( x(t) \), we have \( x'(t) = -0.2e^{-0.2t}(C_1 \cos(3t) + C_2 \sin(3t)) + e^{-0.2t}(-3C_1 \sin(3t) + 3C_2 \cos(3t)) \). At \( t = 0 \), \( x'(0) = -0.2C_1 + 3C_2 = -41 \). Substitute \( C_1 = -20 \) to solve for \( C_2 \): \( -0.2(-20) + 3C_2 = -41 \) gives \( C_2 = -13 \).

Step 9: Write Specific Solution

The specific solution describing the motion is: \[ x(t) = e^{-0.2t}(-20 \cos(3t) - 13 \sin(3t)) \].

Step 10: Determine Damping Nature

The motion is characterized by the nature of the roots of the characteristic equation. Since the discriminant was negative, the roots were complex, which indicates underdamped motion.

Key Concepts

Differential Equation

Damped harmonic motion is often described by a second-order linear differential equation. This equation models the forces acting on a system, such as a mass attached to a spring, in a medium that also exerts a damping force. Mathematically, the differential equation is given by:

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

where:

• \( m \) is the mass of the object,
• \( c \) is the damping coefficient,
• \( k \) is the spring constant,
• \( x \) is the displacement from equilibrium,
• \( \frac{d^2x}{dt^2} \) is the acceleration,
• \( \frac{dx}{dt} \) is the velocity.

This equation shows that the system's acceleration depends on both the damping force and the restoring force from the spring. Finding solutions to this equation helps us understand how the system behaves over time.

Spring Constant

The spring constant \( k \) is a crucial parameter in the differential equation for damped harmonic motion. It represents the stiffness of the spring, or how much force is needed to stretch or compress the spring by a unit distance. A larger spring constant means a stiffer spring, which requires more force to displace.

For our problem, the spring constant is given as \( 226 \) lb/ft. This value is essential because it influences the system's natural frequency and how it responds to external forces. The spring constant appears in the term \( kx \) of the differential equation, where \( x \) is the displacement, indicating that the restoring force is proportional to the displacement.

The larger the spring constant, the more responsive the spring is to changes in displacement, and as such, plays a key role in determining the behavior of the system in motion.

Damping Force

The damping force is the resisting force exerted by the medium in which the system is moving. In damped harmonic motion, this force is usually proportional to the velocity of the moving object, and it works to counteract the motion, eventually bringing it to a stop.

For the given problem, the damping force is specified as being 10 times the instantaneous velocity. Mathematically, this is shown as the term \( c\frac{dx}{dt} \) in the differential equation, where \( c = 10 \) lb·s/ft is the damping coefficient. This indicates that the force resisting the motion increases with higher velocities.

• A high damping coefficient means the system loses energy quickly, slowing the motion sooner.
• A low damping coefficient allows the motion to continue longer before stopping.

Understanding damping forces is crucial, as they determine whether the system will exhibit overdamped, critically damped, or underdamped behavior.

Initial Conditions

Initial conditions are critical for solving differential equations, especially in the context of damped harmonic motion. They define the system's starting state, like its initial position and velocity, which influence how the system evolves over time.

In the exercise, we have two initial conditions:

• \( x(0) = -20 \) ft: The object starts 20 feet below its equilibrium position.
• \( \frac{dx}{dt}(0) = -41 \) ft/s: The object begins with a downward velocity of 41 ft/s.

These initial conditions are used to determine the specific solution of the differential equation. Incorporating them allows us to solve for the constants in the general solution, \( C_1 \) and \( C_2 \), thus providing a complete description of the motion specific to the scenario outlined in the problem.