Problem 33 - Advanced Engineering Mathematics

Question

A 4-lb weight stretches a spring \(2 \mathrm{ft}\). The weight is released from rest 18 in above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to \(\frac{7}{8}\) times the instantaneous velocity. Use the Laplace transform to find the equation of motion \(x(t)\).

Step-by-Step Solution

Answer

The equation of motion is a damped harmonic oscillation described by \(x(t)\), derived from the differential equation and initial conditions, using the Laplace transform.

Step 1: Define the physical system

The problem involves a mass-spring-damper system. The weight is 4 lb, which can be converted to mass using the relation between weight and gravity (where gravity is 32 ft/s²). Hence, the mass \(m = \frac{4}{32} = \frac{1}{8}\) slugs. The spring constant \(k\) is found using Hooke's Law: \(F = kx\). Since a 4-lb force stretches the spring by 2 ft, \(k = \frac{4}{2} = 2\) lb/ft.

Step 2: Establish the damping coefficient

The damping force is proportional to velocity, given by \(\frac{7}{8}v\), where \(v\) is the instantaneous velocity. Thus, the damping coefficient is \(c = \frac{7}{8}\).

Step 3: Set up the differential equation

The motion equation for a damped harmonic oscillator is \( m \ddot{x} + c \dot{x} + kx = 0 \). Plug in the values: \(\frac{1}{8} \ddot{x} + \frac{7}{8} \dot{x} + 2x = 0\). Simplify by multiplying the entire equation by 8: \(\ddot{x} + 7 \dot{x} + 16x = 0\).

Step 4: Apply the Laplace transform

Take the Laplace transform of the entire equation, noting the initial conditions: released from rest means initial velocity \(\dot{x}(0) = 0\), and initial displacement from equilibrium is \( - \frac{18}{12} = -1.5 \) ft. Thus, the Laplace transform of the equation is \( s^2X(s) - sx(0) - \dot{x}(0) + 7[sX(s) - x(0)] + 16X(s) = 0 \).

Step 5: Substitute initial conditions and solve for X(s)

Plug in the initial conditions: \( x(0) = -1.5 \) and \( \dot{x}(0) = 0 \). The equation becomes: \( s^2X(s) + 1.5s + 7(sX(s) + 1.5) + 16X(s) = 0 \). Simplify to get \( (s^2 + 7s + 16)X(s) = -1.5(s + 7) \). Solve for \(X(s)\): \(X(s) = \frac{-1.5(s+7)}{s^2 + 7s + 16}\).

Step 6: Perform partial fraction decomposition

Decompose the expression \( \frac{-1.5(s+7)}{s^2 + 7s + 16} \) into a form suitable for inverse Laplace transformation. The quadratic in the denominator can be factored, or use completing the square to rewrite the terms for known Laplace pairs.

Step 7: Inverse Laplace transform to find x(t)

Using inverse Laplace transforms on the decomposed expression, find \(x(t)\). This typically involves standard forms such as exponential decay combined with oscillatory (sine or cosine) terms.

Step 8: Final form of the equation of motion

Combine results from partial fraction decomposition and inverse Laplace to write the final form for \(x(t)\), which represents the damped harmonic motion of the system.

Key Concepts

Damped Harmonic Oscillator

A damped harmonic oscillator describes a system where a mass is attached to a spring and experiences a damping force as it moves. Imagine a pendulum swinging back and forth; if it encounters air resistance or any friction, its motion will slowly die down. This is what damping means.
In mathematical terms, a damped harmonic oscillator is characterized by a differential equation:

• The mass (m) of the object.
• The damping coefficient (c), representing the strength of damping.
• The spring constant (k), which measures the stiffness of the spring.

The presence of damping modifies the natural oscillations of the system, leading to a decrease in amplitude over time. This reduction in motion can occur in various ways: overdamped, critically damped, and underdamped, based on the value of the damping coefficient.

Mass-Spring-Damper System

A mass-spring-damper system is a fundamental mechanical system used to model oscillations and vibrations. Imagine a car's suspension system cushioning movement; that is how a mass-spring-damper system behaves in real life.
This system consists of three key components:

• The mass (often called inertia), which is represented by the object experiencing forces.
• The spring, providing a restoring force that follows Hooke's Law, where the force is proportional to displacement.
• The damper, which applies a resistance to the motion proportional to its velocity, mimicking frictional forces.

This system's differential equation, as seen in the exercise, combines these elements into a form m\( \ddot{x} + c \dot{x} + kx = 0 \). Solving this equation gives insights into the oscillatory behavior of the system, revealing how quickly it settles down to rest or continues to sway.

Differential Equations

Differential equations play a critical role in modeling physical phenomena, especially in systems involving dynamics and change over time. These equations relate a function with its derivatives, showing how that function changes.
In our context, the differential equation \( m\ddot{x} + c\dot{x} + kx = 0 \) encapsulates the behavior of a damped harmonic oscillator. Each term in this equation corresponds to a physical aspect of the mass-spring-damper system:

• The double dot represents acceleration or the second derivative of the position.
• The single dot denotes velocity or the first derivative of the position.
• The last term signifies the position multiplied by the spring constant, indicating the force from the spring.

Working through these equations helps us predict how the system behaves. Solving them typically involves calculus and sometimes specific techniques like Laplace transforms to manage the complexity.

Inverse Laplace Transform

The Laplace transform is a powerful mathematical tool used to simplify the process of solving differential equations. By transforming functions from the time domain into a complex frequency domain, it converts differential equations into algebraic ones.
The inverse Laplace transform, however, is the crucial step that turns solutions about frequency back to understand them in terms of time. Especially in vibrating systems like the damped harmonic oscillator, applying the inverse Laplace transform to the expression we've processed gives the actual motion over time:

• To get from our algebraic results in 'Laplace-land' back to familiar time-space, inverse transformation is essential.
• It often utilizes pre-calculated forms listed in Laplace transform tables to expedite the process.

Combining both initial conditions and the algebraic solution gives us clear insights into how the system behaves dynamically over time, neatly connecting theory with the tangible motion we can observe.