Problem 93 - Calculus 3

Question

An 800-lb weight (25 slugs) is attached to a vertical spring with a spring constant of 226 lb/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 ft below its equilibrium position with a downward velocity of 41 ft/sec. b. Graph the solution and determine whether the motion is over damped, critically damped, or under damped.

Step-by-Step Solution

Answer

The motion is underdamped with equation of motion: \( x(t) = e^{-0.2t} (-20\cos(3t) + 13.667\sin(3t)) \).

Step 1: Write the Differential Equation

The motion of the mass-spring-damper system can be described by the second order linear differential equation: \[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \] where \( m = 25 \) slugs, \( c = 10 \), and \( k = 226 \) lb/ft.

Step 2: Apply Initial Conditions

Given the initial conditions: the weight is released from a position 20 ft below equilibrium (\( x(0) = -20 \)) and has a downward velocity of 41 ft/sec (\( x'(0) = 41 \)), these conditions will be used to solve for constants in the differential equation's general solution.

Step 3: Determine the Characteristic Equation

The characteristic equation for the system can be written as: \[ 25r^2 + 10r + 226 = 0 \]. Solving this quadratic equation will help establish the nature of the damping.

Step 4: Solve the Characteristic Equation

Solve the quadratic equation \( 25r^2 + 10r + 226 = 0 \) using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substituting \( a = 25 \), \( b = 10 \), and \( c = 226 \), calculate the discriminant \( b^2 - 4ac \) to determine the roots.

Step 5: Analyze the Discriminant

Calculate the discriminant: \( 10^2 - 4 \times 25 \times 226 = -2250 \). Since the discriminant is negative, the roots are complex, indicating the system is underdamped. The roots are \( r = \frac{-10 \pm i\sqrt{2250}}{50} \).

Step 6: Write the General Solution

Since the roots are complex, the general solution of the differential equation is: \( x(t) = e^{-0.2t} (c_1 \cos(3t) + c_2 \sin(3t)) \).

Step 7: Apply Initial Conditions to the Solution

Substitute the initial conditions \( x(0) = -20 \) and \( x'(0) = 41 \) into the general solution to solve for \( c_1 \) and \( c_2 \).

Step 8: Solve for \( c_1 \) and \( c_2 \)

Using the initial condition \( x(0) = -20 \), \( -20 = e^0(c_1) \) gives \( c_1 = -20 \). Differentiate the general solution and substitute \( x'(0) = 41 \) to solve for \( c_2 \).

Step 9: Find the Final Solution

The final equation of motion after determining \( c_1 \) and \( c_2 \) is: \( x(t) = e^{-0.2t} (-20 \cos(3t) + 13.667 \sin(3t)) \).

Step 10: Graph and Determine Damping

Graph the function \( x(t) = e^{-0.2t} (-20 \cos(3t) + 13.667 \sin(3t)) \), observing the oscillatory behavior that dampens over time, confirming it as underdamped.

Key Concepts

Damping Force

When we talk about a damping force, we're looking at a force that acts to slow down motion in a system. Imagine a mass hanging on a spring. It bounces up and down when disturbed, but what if there is some resistance, like a thick fluid like honey? This resistance is the damping force. In this particular problem, the damping force is proportional to the velocity of the mass.
It is essentially 10 times the speed of the mass, which helps prevent the weight from oscillating endlessly.

• Proportional relationship: The damping force is directly proportional to the velocity, expressed mathematically as \( F_d = c \cdot v \) where \( c \) is the damping coefficient and \( v \) is the velocity.
• Damping coefficient: In our problem, \( c = 10 \) suggests a moderate level of damping which opposes the motion of the mass.
• Effects of damping: With positive damping, the amplitude of oscillations will decrease over time, leading to a type of motion called underdamping, as found by solving the characteristic equation.

By considering the damping force, we can predict how quickly the oscillations diminish, providing us with valuable insights into the behavior of real-world systems.

Mass-Spring-Damper System

A mass-spring-damper system is an essential concept when studying mechanical systems affecting how objects move. It combines a mass (weight), a spring component, and a damping force (resistance), allowing us to describe this system with a second-order differential equation. The system involves:

• Mass (\(m\)): Here, the mass is given as 25 slugs, representing the weight converted using gravitational forces.
• Spring Constant (\(k\)): This spring has a constant of 226 lb/ft, describing the stiffness of the spring. A higher value indicates a stiffer spring.
• Damping Force (\(c\)): As discussed earlier, this resistance factor is 10 times the velocity in this scenario.

Using these elements, we derive the equation of motion using the standardized formula:
\[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \]
By solving this equation, we determine how the mass behaves with time, providing a critical framework to solving real-world engineering challenges.

Initial Conditions

When solving differential equations for problems like a mass-spring-damper system, initial conditions are crucial. They provide specific values at the start of observation to tailor the general solution to a particular scenario. In the given example, two initial conditions are given:

• Initial Position (\(x(0) = -20\)): This indicates that the system starts 20 feet below its equilibrium position. Equilibrium is where the spring force perfectly balances the weight.
• Initial Velocity (\(x'(0) = 41\) ft/sec): This describes the speed at which the system initially moves downward.

Applying these conditions allows us to solve for specific constants within the general solution of the differential equation. These constants adjust the solution to reflect the real behavior of the system based on how it begins. For instance, using our conditions, we solved for constants \(c_1\) and \(c_2\) which personalize the oscillatory equation of motion expressing the system's response over time. Understanding initial conditions makes the abstract math of second-order differential equations applicable and meaningful in practical scenarios.