Problem 22 - Advanced Engineering Mathematics

Question

A 4 -foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers damping force numerically equal to \(\sqrt{2}\) times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of \(5 \mathrm{ft} / \mathrm{s}\). Find the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?

Step-by-Step Solution

Answer

The extreme displacement occurs at \( t = \frac{\pi}{4} \) and the position is \(-\frac{5}{2} \cdot e^{\sqrt{2}\pi/2}\).

Step 1 - Determine Spring Constant

The spring stretches from 4 ft to 8 ft under the influence of an 8 lb weight. The force due to weights is given by the formula: \[ F = k imes x \] where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement. Given \( F = 8 \) lbs and \( x = 4 \) ft, we can solve for the spring constant \( k \):\[ k = \frac{8}{4} = 2 \, \text{lb/ft} \].

Step 2 - Write the Differential Equation of Motion

The mass-spring-damper system gives us the differential equation:\[ m \frac{d^2y}{dt^2} + c \frac{dy}{dt} + ky = 0 \]Substitute the mass \( m \), damping coefficient \( c = \sqrt{2} \), and spring constant \( k = 2 \). The mass can be found from \( m = \frac{W}{g} = \frac{8}{32} = \frac{1}{4} \) slugs (where \( g = 32 \text{ ft/s}^2 \)), resulting in the equation:\[ \frac{1}{4} \frac{d^2y}{dt^2} + \sqrt{2} \frac{dy}{dt} + 2y = 0\].

Step 3 - Simplify and Solve the Differential Equation

Rearrange and simplify the equation by multiplying through by 4:\[ \frac{d^2y}{dt^2} + 4\sqrt{2} \frac{dy}{dt} + 8y = 0 \]The characteristic equation is:\[ r^2 + 4\sqrt{2}r + 8 = 0 \]Solve using the quadratic formula:\[ r = \frac{-4\sqrt{2} \pm \sqrt{(4\sqrt{2})^2 - 4 \cdot 8}}{2} = -2\sqrt{2} \pm 2 \], yielding roots \( r_1 = - (2\sqrt{2} + 2) \) and \( r_2 = - (2\sqrt{2} - 2) \).

Step 4 - Formulate General Solution

With complex roots, the general solution of the differential equation is:\[ y(t) = e^{-2\sqrt{2}t} (C_1 \cos 2t + C_2 \sin 2t) \].

Step 5 - Apply Initial Conditions

The initial conditions are given as initial position \( y(0) = 0 \) and initial velocity \( y'(0) = -5 \), since the mass is released from equilibrium with a downward velocity. Apply \( y(0) = 0 \):\[ y(0) = e^0(C_1 \cos 0 + C_2 \sin 0) = 0 \Rightarrow C_1 = 0 \]The derivative of the solution is:\[ y'(t) = -2\sqrt{2} e^{-2\sqrt{2}t}(C_1 \cos 2t + C_2 \sin 2t) + e^{-2\sqrt{2}t}(C_2 \cos 2t - 2C_1 \sin 2t) \]Using \( y'(0) = -5 \):\[ y'(0) = 2C_2 = -5 \Rightarrow C_2 = -\frac{5}{2} \]The solution is:\[ y(t) = e^{-2\sqrt{2}t} \left(-\frac{5}{2} \sin 2t\right) \].

Step 6 - Find Time of Extreme Displacement

Extreme displacement occurs when the velocity \( y'(t) = 0 \):\[ y'(t) = e^{-2\sqrt{2}t}(-\frac{5}{2} \times 2\cos 2t - (-\frac{5}{2} \sin 2t)2\sqrt{2}) = 0 \]Simplify and solve for \( t \) when the derivative equals zero. This implies finding \( \cos 2t = 0 \), solving gives:\[ 2t = \frac{\pi}{2} + n\pi, \, n \in \mathbb{Z} \rightarrow t = \frac{\pi}{4} + \frac{n\pi}{2} \].

Step 7 - Calculate Position at Extreme Displacement Time

Use the smallest positive value of \( t = \frac{\pi}{4} \) to determine position:\[ y\left(\frac{\pi}{4}\right) = e^{-2\sqrt{2}\frac{\pi}{4}} \left(-\frac{5}{2}\sin \frac{\pi}{2}\right) \]Calculate \( y\left(\frac{\pi}{4}\right) = -\frac{5}{2}\cdot e^{\sqrt{2}\pi/2} \).

Key Concepts

Spring Constant

The spring constant, denoted as \( k \), is a crucial factor that quantifies a spring's stiffness. In simple terms, it measures the amount of force needed to stretch or compress a spring by a unit length. The higher the spring constant, the stiffer the spring is. It is derived using Hooke's Law, which is given by the formula:
\[ F = k imes x \]where:

• \( F \) is the force exerted on the spring,
• \( k \) is the spring constant,
• \( x \) is the displacement of the spring from its rest position.

In the given problem, a force of 8 pounds causes an 8-foot spring to stretch to 4 feet beyond its natural length, making the displacement \( x = 4 \) feet. Solving for \( k \), we have:
\[ k = \frac{8}{4} = 2 \, \text{lb/ft} \]Thus, the spring constant is 2 lb/ft, indicating the amount of force needed per foot to stretch or compress the spring.

Damping Force

Damping force is a resisting force that acts opposite to the direction of motion, usually proportional to the velocity of the object moving through a medium. It is introduced into a system to reduce vibrations and oscillations. The damping force is critical in bringing a vibrating system to rest in an efficient manner.
In the mass-spring-damper system, the damping force is described as:
\[ F_d = c \times v \]where:

• \( F_d \) is the damping force,
• \( c \) is the damping coefficient,
• \( v \) is the instantaneous velocity of the moving mass.

In this scenario, the damping coefficient \( c \) is \( \sqrt{2} \), signifying that the damping force is \( \sqrt{2} \) times the velocity. This value helps in counteracting the energy in the system, eventually leading the mass to return to its equilibrium position smoothly.

Mass-Spring-Damper System

The mass-spring-damper system is a classical model used in physics and engineering to explain and analyze the motion of a mass attached to a spring with a damping component. This model is used to represent the balance of forces acting on the mass.
The differential equation that governs this system is as follows:
\[ m \frac{d^2y}{dt^2} + c \frac{dy}{dt} + ky = 0 \]where:

• \( m \) is the mass,
• \( c \) is the damping coefficient,
• \( k \) is the spring constant,
• \( y \) is the displacement from the equilibrium position.

This equation takes into account the forces due to the spring, damping, and mass. It is a second-order linear differential equation that describes how the mass moves over time, responding to forces from the spring and damper.
In the problem, the mass calculation yields \( m = \frac{1}{4} \) slugs. Combining this with our other known values, the equation becomes simplified to:
\[ \frac{d^2y}{dt^2} + 4\sqrt{2} \frac{dy}{dt} + 8y = 0 \]This form reflects how the particular system will respond over time, with its damping and spring constant influencing the nature of the motion.

Initial Conditions

Initial conditions are critical parameters in solving differential equations, defining the system's state at the beginning of observation, typically at time \( t = 0 \). For dynamic systems like the mass-spring-damper system, they specify the object's initial displacement and velocity.
In this case study, we have:

• Initial position \( y(0) = 0 \)
• Initial velocity \( y'(0) = -5 \, \text{ft/s} \)

These initial conditions are used to determine specific constants in the general solution of the differential equation. They allow the formulation of a solution that precisely matches the actual behavior of the system right from the start.
Applying these conditions yields the constants \( C_1 = 0 \) and \( C_2 = -\frac{5}{2} \), leading to a particular solution:
\[ y(t) = e^{-2\sqrt{2}t} \left(-\frac{5}{2} \sin 2t\right) \] This solution reflects the system's dynamics with its initial velocity and position integrated into the expression for \( y(t) \).