Problem 96 - Calculus Volume 3

Question

A 32-lb weight (1 slug) stretches a vertical spring 4 in. The resistance in the spring-mass system is equal to four times the instantaneous velocity of the mass. a. Find the equation of motion if it is released from its equilibrium position with a downward velocity of \(12 \mathrm{ft} / \mathrm{sec}\). b. Determine whether the motion is overdamped, critically damped, or underdamped.

Step-by-Step Solution

Answer

The equation of motion is \(x(t) = 2e^{-2t} \sin(6t)\); the motion is underdamped.

Step 1: Convert Units and Define Model

First, we need to convert the given units and identify key parameters. The weight of the mass is 32 pounds, and since 1 slug weighs 32 pounds on Earth, the mass (\(m\)) is 1 slug. The spring is stretched 4 inches, which is equivalent to \(\frac{1}{3}\) feet (since 1 foot = 12 inches). The displacement from equilibrium produces a force of weight, so using Hooke’s Law, \(mg = kx\), where \(k\) is the spring constant. This gives \(32 = k \times \frac{1}{3}\), hence \(k = 96\) lb/ft. The damping coefficient \(c\) is equal to four times the velocity, so \(c = 4\).

Step 2: Formulate the Differential Equation

The equation of motion for a damped spring-mass system is given by the second-order linear differential equation: \(m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0\). By substituting \(m = 1\), \(c = 4\), and \(k = 96\) from Step 1, we get: \(\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 96x = 0\).

Step 3: Apply Initial Conditions

The mass is released from its equilibrium position with a downward velocity, which means its initial displacement \(x(0) = 0\) and initial velocity \(\frac{dx}{dt}(0) = -12\) ft/s (since downward is typically negative). These will serve as initial conditions for solving the differential equation.

Step 4: Solve the Characteristic Equation

The characteristic equation of the differential equation \(\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 96x = 0\) is \(r^2 + 4r + 96 = 0\). Solving using the quadratic formula \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we find \(r = \frac{-4 \pm \sqrt{16 - 384}}{2}\). The discriminant \(b^2 - 4ac = 16 - 384 = -368\), which is less than zero, indicating complex roots.

Step 5: Write the General Solution

Since the roots are complex \(r = -2 \pm 6i\), the general solution of the differential equation is \(x(t) = e^{-2t}(C_1 \cos(6t) + C_2 \sin(6t))\).

Step 6: Apply Initial Conditions to General Solution

Using the initial conditions: \(x(0) = 0\), gives \(C_1 = 0\) since \(e^0 = 1\) and \(\cos(0) = 1\), \(\sin(0) = 0\). The derivative \(x'(t) = e^{-2t}(-2C_2 \sin(6t) + 6C_2 \cos(6t))\). Applying \(x'(0) = -12\), we get \(-2C_2 = -12\), hence \(C_2 = 2\).

Step 7: Conclude the Equation of Motion

Substituting \(C_1 = 0\) and \(C_2 = 2\) into the general solution, the equation of motion is \(x(t) = 2e^{-2t} \sin(6t)\).

Step 8: Classify the Damping

Given that the discriminant \(b^2 - 4ac = -368\) was negative, leading to complex roots, the system is underdamped. Underdamped systems have oscillatory motion.

Key Concepts

Differential Equations

Differential equations are crucial in modeling physical systems, and the spring-mass system is a classic example. They describe how a quantity changes over time, showing the dynamics of systems based on their environment. In the context of a spring-mass system, the differential equation involves displacement, velocity, and acceleration of the mass. For this problem, the differential equation is second-order, meaning it involves the second derivative of displacement. These types of equations are typically expressed as

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

Here, \( m \) is the mass, \( c \) is the damping coefficient, and \( k \) is the spring constant. Each component of the equation reflects a force acting on the system:

• The acceleration term, \( \frac{d^2x}{dt^2} \), relates to inertia,
• The term \( c \frac{dx}{dt} \) represents damping, slowing the system,
• And \( kx \) depicts the restoring force of the spring.

By understanding these components, students can see how differential equations model real-world phenomena by capturing the interplay of physical forces.

Damping Systems

A damping system features a force that works against the motion, eventually bringing a moving object to rest. In mechanical systems like our spring-mass example, damping is the resistance proportional to the velocity. When damping is introduced, it significantly alters the system's characteristics. There are three primary types of damping:

• **Overdamping:** Characterized by real and distinct roots in the characteristic equation, leading to a slow return without oscillation,
• **Critical Damping:** Occurs when the system returns to equilibrium as quickly as possible without oscillating,
• **Underdamping:** Identified by complex roots, this results in oscillations that decrease over time due to damping.

In this exercise, the negative discriminant of the characteristic equation indicates underdamping. This suggests that the spring-mass system will exhibit oscillatory motion but with decreasing amplitude over time.

Initial Conditions

Initial conditions are essential for solving differential equations, allowing us to find specific solutions that match the scenario. These conditions define the system's state at a starting point, usually time \( t=0 \). For our spring-mass system, initial conditions involve the initial displacement and velocity of the mass. They dictate how the system will behave from the starting moment, influencing the constants within the general solution of the differential equation.

For the given exercise, the initial conditions were:

• The mass is at the equilibrium position, denoting initial displacement, \( x(0) = 0 \),
• A downward initial velocity of \( -12 \frac{ft}{s} \), becoming \( \frac{dx}{dt}(0) = -12 \).

Applying these initial conditions helps derive specific constants in the general solution. They ensure the model accurately represents how the system begins its motion, influencing the resulting motion pattern.