Problem 95 - Calculus Volume 3

Question

A 1 -kg mass stretches a spring \(6.25 \mathrm{~cm}\). The resistance in the spring-mass system is equal to eight times the instantaneous velocity of the mass. a. Find the equation of motion if the mass is released from a position \(5 \mathrm{~m}\) below its equilibrium position with an upward velocity of \(10 \mathrm{~m} / \mathrm{sec}\). b. Determine whether the motion is overdamped, critically damped, or underdamped.

Step-by-Step Solution

Answer

The equation of motion is \( x(t) = e^{-4t}(-5\cos(\sqrt{92.8}t) + \frac{10}{\sqrt{92.8}}\sin(\sqrt{92.8}t)) \), and the system is underdamped.

Step 1: Understanding the System

We begin by understanding the spring-mass system and resistive force. The spring force can be represented as: \[ F_s = -kx \] where \( k \) is the spring constant and \( x \) is the displacement from equilibrium. The damping force is given by:\[ F_d = -8v \] with \( v \) being the velocity. Using Hooke's Law, which states that the force needed to extend the spring is proportional to the distance it stretches by: \[ k x = mg \]Given: mass \( m = 1 \text{ kg} \), stretched distance \( x = 6.25 \text{ cm} = 0.0625 \text{ m} \). Also, weight \( mg = kx \) thus, the acceleration due to gravity \( g = 9.8 \text{ m/s}^2 \). Now solve for \( k \):\[ k = \frac{mg}{x} = \frac{1 \times 9.8}{0.0625} = 156.8 \text{ N/m} \].

Step 2: Formulating the Equation of Motion

The differential equation for a damped harmonic oscillator is:\[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]where \( c = 8m \) (since the resistance is 8 times the velocity), leading to \( c = 8 \times 1 = 8 \). Thus the equation becomes:\[ \frac{d^2x}{dt^2} + 8 \frac{dx}{dt} + 156.8x = 0 \].

Step 3: Initial Conditions

To find the specific solution, we need initial conditions. Given that the mass is released from a position 5 meters below the equilibrium, let the initial position \( x(0) = -5 \). The initial velocity \( \frac{dx}{dt}(0) = 10 \text{ m/s upward} \succ (positive) \). These will be used to solve the subsequent differential equation.

Step 4: Solving the Characteristic Equation

The characteristic equation for this second-order linear homogeneous differential equation is:\[ r^2 + 8r + 156.8 = 0 \].Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) gives:\[ r = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times 156.8}}{2} = -4 \pm \sqrt{-92.8} \].The discriminant \( -92.8 \) is negative, implying complex roots.

Step 5: Determining Damping

The roots are complex, given by \( r = -4 \pm \sqrt{-92.8} \), indicating the system is underdamped. The general solution is:\[ x(t) = e^{-4t}(C_1 \cos(\sqrt{92.8} \cdot t) + C_2 \sin(\sqrt{92.8} \cdot t)) \].

Step 6: Applying the Initial Conditions

Use the initial conditions to solve for \( C_1 \) and \( C_2 \). At \( t = 0 \), \( x(0) = -5 \) gives:\[ x(0) = C_1 = -5 \].\( \frac{dx}{dt}(0) = 10 \) and using the derivative of the function:\[ x'(t) = e^{-4t}((-4C_1 - \sqrt{92.8}C_2) \cdot \cos(\sqrt{92.8}t) + (\sqrt{92.8}C_1 - 4C_2) \cdot \sin(\sqrt{92.8}t)) \].At \( t=0 \), this simplifies to \( -4(-5) - \sqrt{92.8}C_2 = 10 \), solving gives \( C_2 \).

Step 7: Solving for C2

Substitute \( C_1 = -5 \) into the solved equation for initial velocity, solve for \( C_2 \):\[ -4(-5) - \sqrt{92.8}C_2 = 10 \]\[ 20 - \sqrt{92.8}C_2 = 10 \]\[ C_2 = \frac{10}{\sqrt{92.8}} \].

Step 8: Final Equation of Motion

Substitute back \( C_1 \) and \( C_2 \) values:\[ x(t) = e^{-4t}(-5\cos(\sqrt{92.8}t) + \frac{10}{\sqrt{92.8}}\sin(\sqrt{92.8}t)) \].

Step 9: Conclusion: Type of Damping

The motion is "underdamped" based on the complex roots of the characteristic equation.

Key Concepts

Equation of Motion

In the study of damped harmonic oscillators, the equation of motion is a crucial component. It represents how the spring-mass system moves over time. The general form for a damped harmonic oscillator is \[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]Here, \( m \) is the mass of the system, \( c \) is the damping coefficient, and \( k \) is the spring constant. This differential equation combines the effects of inertia, damping, and spring force.
In our exercise, the equation simplifies to\[ \frac{d^2x}{dt^2} + 8 \frac{dx}{dt} + 156.8x = 0 \]This form shows a relationship between acceleration, velocity, and position. Solving this equation helps predict future motion of the mass.

Spring Constant

The spring constant \( k \) describes how stiff or flexible a spring is. It’s determined by how much force is needed to stretch or compress the spring by a certain distance. Hooke's Law, \( F_s = -kx \), governs this behavior.
In our case, the spring stretches by \(6.25\) cm due to the weight of a 1-kg mass. Using the equation \( kx = mg \), where \( g \) is the gravitational acceleration (\(9.8 \text{ m/s}^2\)), we can solve for \( k \): \[ k = \frac{mg}{x} = \frac{1 \times 9.8}{0.0625} = 156.8 \text{ N/m} \]This value tells us how resistant the spring is to deformation when a force is applied.

Differential Equation

Differential equations are mathematical equations that relate a function to its derivatives. In the context of harmonic oscillators, they describe how the system evolves over time due to forces acting upon it. For a damped spring-mass system, the differential equation is \[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]where \( m \), \( c \), and \( k \) represent mass, damping coefficient, and spring constant respectively.
By solving this equation, we can understand the system's behavior under various conditions. In our problem, we ended up with a differential equation that has to be solved to understand the damped motion of the mass involved.

Damping Types

Damping is an essential concept in harmonic oscillators, as it determines how quickly the motion ceases. There are three types of damping:

• Overdamped - motion returns to equilibrium without oscillation.
• Critically Damped - motion returns to equilibrium in the shortest time without oscillating.
• Underdamped - oscillates with gradually decreasing amplitude.

In our solution, the characteristic equation yielded complex roots, indicating an underdamped system. This means the mass oscillates while gradually coming to a stop. Understanding the type of damping helps in predicting how quickly the system's motion will taper off.

Initial Conditions

Initial conditions are vital for solving the differential equation of a system. They specify the status of the system at the beginning of the problem. In the discussed problem, the initial conditions were:

• Initial Position - The mass is released 5 meters below the equilibrium position, therefore \( x(0) = -5 \).
• Initial Velocity - The mass was given an upward velocity of 10 m/s, making \( \frac{dx}{dt}(0) = 10 \text{ m/s} \).

By applying these conditions, we can find the unique solution for our differential equation. This step is crucial for predicting the future motion of the spring-mass system accurately.