Problem 23 - Advanced Engineering Mathematics

Question

A 1-kilogram mass is attached to a spring whose constant is \(16 \mathrm{~N} / \mathrm{m}\), and the entire system is then submerged in a liquid that imparts a darnping force numerically equal to 10 times the instantaneous velocity. Determine the equations of motion if (a) the mass is initially released from rest from a point 1 meter below the equilibrium position, and then (b) the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of \(12 \mathrm{~m} / \mathrm{s}\).

Step-by-Step Solution

Answer

(a) \( x(t) = -0.25e^{-2t} - 0.75e^{-8t} \); (b) \( x(t) = 7e^{-2t} - 8e^{-8t} \).

Step 1: Identify System Dynamics

In this mechanical system, we have a mass-spring-damper setup, where the displacement from equilibrium is referred to as \( x(t) \). Here, the mass \( m = 1 \) kg, the spring constant \( k = 16 \) N/m, and the damping coefficient \( c = 10 \). The differential equation governing this motion is derived from Newton's second law, giving us \( m x'' + c x' + k x = 0 \). Substituting the values, we get the characteristic equation: \( x'' + 10x' + 16x = 0 \).

Step 2: Solve the Characteristic Equation

The characteristic equation from the differential equation is \( r^2 + 10r + 16 = 0 \). To find the roots, use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 10 \), and \( c = 16 \). This gives us \( r = \frac{-10 \pm \sqrt{100 - 64}}{2} = \frac{-10 \pm 6}{2} \), leading to roots \( r_1 = -2 \) and \( r_2 = -8 \).

Step 3: Formulate General Solution

Since the roots \( r_1 = -2 \) and \( r_2 = -8 \) are real and distinct, the general solution is \( x(t) = C_1 e^{-2t} + C_2 e^{-8t} \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.

Step 4: Apply Initial Condition (a) Release from Rest

For case (a), the initial conditions provided are \( x(0) = -1 \) (starting 1 meter below equilibrium) and \( x'(0) = 0 \) (released from rest). Substitute these into the general solution and its derivative: \( C_1 + C_2 = -1 \) and \(-2C_1 - 8C_2 = 0\). Solving these equations gives \( C_1 = -0.25 \) and \( C_2 = -0.75 \). Therefore, the equation of motion is \( x(t) = -0.25e^{-2t} - 0.75e^{-8t} \).

Step 5: Apply Initial Condition (b) Release with Velocity

For case (b), the initial conditions are \( x(0) = -1 \) and \( x'(0) = 12 \). Substituting these, we get \( C_1 + C_2 = -1 \) and \( -2C_1 - 8C_2 = 12 \). Solving this system, we find \( C_1 = 7 \) and \( C_2 = -8 \). Thus, the equation of motion is \( x(t) = 7e^{-2t} - 8e^{-8t} \).

Key Concepts

Differential Equations

Differential equations are a fundamental concept in understanding mechanical vibrations. They are mathematical equations that relate a function to its derivatives, representing how a system changes over time. In the context of a vibrating system like a mass-spring-damper, differential equations describe the motion of the mass as a function of time.
In our example, the differential equation is derived from Newton’s second law of motion, which states that the sum of forces acting on an object is equal to the mass of the object multiplied by its acceleration. This gives us the equation:

• The left side: combining mass, damping, and elastic forces, written as: \( m x'' + c x' + k x \)
• The right side: typically set to zero when analyzing free vibrations (no external forces).

Substituting known values in gives us a specific differential equation for this problem. Understanding how to set up and solve these equations is key for predicting system behavior.

Mass-Spring-Damper System

The mass-spring-damper system is a standard model used in mechanics to describe real-world oscillatory systems like car suspensions or building structures facing earthquakes. Here's a closer look at each component:

• **Mass (m)**: Represents the object that is moving. In our problem, it's a 1-kilogram mass.
• **Spring (k)**: Provides a restoring force that tries to bring the mass back to the equilibrium position. The spring constant in this case is \( 16 \, \mathrm{N/m} \), indicating stiffness.
• **Damper (c)**: Opposes motion through resistance, similar to friction. Here, the damper force depends on velocity and is characterized by a damping coefficient, set as \( 10 \).

Together, these components form the basis of our system’s dynamics, determining how the mass moves over time. If either the spring constant or damping is significantly changed, the system’s behavior will differ.

Characteristic Equation

The characteristic equation is a crucial step in solving differential equations for mechanical vibrations. By finding the roots of this polynomial equation, we can determine the type of motion the system exhibits. For the differential equation \( x'' + 10x' + 16x = 0 \), the characteristic equation is derived by replacing each derivative with powers of \( r \), resulting in \( r^2 + 10r + 16 = 0 \).
To solve this:

• Use the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where\( a = 1 \), \( b = 10 \), \( c = 16 \).
• This leads to roots \( r_1 = -2 \) and \( r_2 = -8 \), which are real and distinct.

These roots indicate that the system is overdamped. In this case, the mass will return to equilibrium without oscillating, but it does so at different rates due to the distinct roots.

Initial Conditions

Initial conditions are the starting parameters for a dynamic system's motion, crucial for determining the specific solution to the differential equations. In our scenarios, two sets of initial conditions are provided:

• **Case (a):** The mass is released from rest at 1 meter below the equilibrium. This means \( x(0) = -1 \) and \( x'(0) = 0 \).
• **Case (b):** The mass starts 1 meter below equilibrium but with an upward velocity of 12 m/s, giving \( x(0) = -1 \) and \( x'(0) = 12 \).

By applying these conditions to the general solution \( x(t) = C_1 e^{-2t} + C_2 e^{-8t} \), we can find the specific constants \( C_1 \) and \( C_2 \). These values tailor the general motion equation to match the initial circumstances of each problem, providing insight into the dynamics just after the system is set into motion.