Problem 31 - Advanced Engineering Mathematics

Question

A mass of 1 slug, when attached to a spring, stretches it 2 feet and then comes to rest in the equilibrium position. Starting at \(t=0\), an external force equal to \(f(t)=8 \sin 4 t\) is applied to the system. Find the equation of motion if the surrounding medium offers a damping force numerically equal to eight times the instantaneous velocity.

Step-by-Step Solution

Answer

The equation of motion is \( x(t) = \frac{1}{5} e^{-4t} - \frac{1}{10} \sin 4t - \frac{1}{5} \cos 4t \).

Step 1: Identify the known parameters and initial conditions

We have a mass of 1 slug attached to a spring and a damping force proportional to the velocity. The spring constant can be determined from the fact that the mass stretches the spring by 2 feet in equilibrium. The damping force is given as 8 times the velocity of the mass. We also have an external force: \(f(t) = 8 \sin 4t\). At \(t=0\), the system starts at rest.

Step 2: Determine the spring constant

We use Hooke's law, \( F = kx \), where \( F \) is the force exerted by the mass (equal to its weight, 1 slug × 32 ft/s²), and \( x \) is the displacement (2 feet). \( k \) is the spring constant.\[ k \times 2 = 32 \quad \Rightarrow \quad k = 16 \text{ lb/ft} \]

Step 3: Set up the differential equation of motion

The damped forced motion of the system with mass \( m \), damping constant \( c \), spring constant \( k \), and external force \( f(t) \) is described by:\[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = f(t) \]Substituting \( m = 1 \), \( c = 8 \), \( k = 16 \), and \( f(t) = 8 \sin 4t \):\[ \frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 16x = 8 \sin 4t \]

Step 4: Solve the homogeneous equation

The homogeneous equation is:\[ \frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 16x = 0 \]The characteristic equation is:\[ r^2 + 8r + 16 = 0 \]Solving for \( r \):\[ r = \frac{-8 \pm \sqrt{8^2 - 4 \cdot 16}}{2} = \frac{-8 \pm 0}{2} = -4 \]So, the solution is \( x_h(t) = C_1 e^{-4t} + C_2 t e^{-4t} \).

Step 5: Solve the particular solution

We look for a particular solution of the form \( x_p(t) = A \sin 4t + B \cos 4t \). Differentiating and substituting into the differential equation to find \( A \) and \( B \):\( -16A \sin 4t - 16B \cos 4t + 32A \cos 4t - 32B \sin 4t + 16(A \sin 4t + B \cos 4t) = 8 \sin 4t \).Equating coefficients, solve: \[ -16A - 32B = 8 \quad \text{and} \quad -32A + 16B = 0 \]

Step 6: Solve the system of equations

Solving:\[ -16A - 32B = 8 \quad (1) \]\[ -32A + 16B = 0 \quad (2) \]From (2): \( B = 2A \), substituting into (1):\[ -16A - 32(2A) = 8 \quad \Rightarrow \quad -80A = 8 \quad \Rightarrow \quad A = -\frac{1}{10} \]\[ B = 2(-\frac{1}{10}) \quad \Rightarrow \quad B = -\frac{1}{5} \]

Step 7: Construct the general solution

The general solution is the sum of the homogeneous and particular solutions:\[ x(t) = C_1 e^{-4t} + C_2 t e^{-4t} - \frac{1}{10} \sin 4t - \frac{1}{5} \cos 4t \]

Step 8: Apply initial conditions

Given the system is at rest at equilibrium at \( t = 0 \), \( x(0) = 0 \) and \( \frac{dx}{dt}(0) = 0 \), substitute to find constants \( C_1 \) and \( C_2 \). From \( x(0) = 0 \):\[ C_1 - \frac{1}{5} = 0 \quad \Rightarrow \quad C_1 = \frac{1}{5} \]From \( \frac{dx}{dt}(0) = 0 \):\[ -4C_1 - \frac{1}{10 (4)} + 4B = 0 \quad \Rightarrow \quad C_2 = 0 \]

Step 9: Finalize the equation of motion

Substituting \( C_1 \) and \( C_2 \) back into the general solution, the equation of motion is:\[ x(t) = \frac{1}{5} e^{-4t} - \frac{1}{10} \sin 4t - \frac{1}{5} \cos 4t \]

Key Concepts

Damped Harmonic Motion

Damped harmonic motion is a type of motion where a system experiences a resistance that opposes its movement. This resistance, known as damping, gradually decreases the amplitude of oscillation over time. Consider a spring-mass system: when a mass is displaced from its equilibrium position and released, it oscillates back and forth.
This motion can be described by a differential equation. However, in damped harmonic motion, an additional term representing the damping force is included, typically proportional to the velocity of the mass. In the case of our problem, the damping force is described as eight times the instantaneous velocity. This creates a term in the differential equation as follows:

• \( c \frac{dx}{dt} \)
• where \( c \) is the damping constant
• \( \frac{dx}{dt} \) is the velocity

This means the system gradually comes to rest as the energy is dissipated due to the damping force.

Spring-Mass System

A spring-mass system is a classic example used to explore dynamics in physics, particularly important in the study of harmonic motion. It consists of a mass attached to the end of a spring that is assumed to be perfectly elastic. When the mass is displaced from its equilibrium position, it exerts a force on the spring, causing it to either compress or stretch.
According to Hooke's Law, the force exerted by a spring is proportional to the displacement of the spring, given by the equation: \[ F = kx \] Where:

• \( F \) is the force exerted by the spring
• \( k \) is the spring constant
• \( x \) is the displacement

A spring-mass system without any external force or damping would oscillate indefinitely. However, in reality, there is often damping, leading to diminished oscillation over time, as observed in the problem where the system comes to rest.

Forced Vibrations

Forced vibrations occur when an external force acts on an oscillating system. This force can be periodic or random, but it tends to sustain the oscillation or even increase its amplitude, depending on the force's frequency in relation to the system's natural frequency.
In the exercise, a force defined by \( f(t) = 8 \sin 4t \) is applied, which suggests a continuous, sinusoidal external input at a specific frequency. In forced vibration scenarios:

• If the frequency of the external force matches the natural frequency of the system, resonance may occur, leading to large oscillations.
• The system's overall response is a combination of the natural damped response and the forced response.

In our differential equation, the term \( f(t) \) represents this continuous "driving" force applied to the system, adding complexity to its motion.

Initial Conditions

Initial conditions are crucial in solving differential equations, as they provide specific information at the start of the motion that determines the constants of integration. They answer the question: where do we begin at \( t = 0 \)?
In the context of this problem, the system starts from rest at the equilibrium position, meaning:

• The displacement of the mass, \( x(0) \), is 0.
• The initial velocity, \( \frac{dx}{dt}(0) \), is also 0.

These conditions allow us to find the constants \( C_1 \) and \( C_2 \) in our solution to the differential equation. By specifically implementing these values, we narrow down the broad set of possible solutions to one that's unique to this situation, fitting the physical scenario perfectly. Initial conditions thus ensure our solution is applicable to the real-world problem at hand.