Problem 12 - Advanced Engineering Mathematics

Question

A mass of 1 slug is suspended from a spring whose spring constant is \(9 \mathrm{lb} / \mathrm{ft}\). The mass is initially released from a point 1 foot above the equilibrium position with an upward velocity of \(\sqrt{3} \mathrm{ft} / \mathrm{s}\). Find the times for which the mass is heading downward at a velocity of \(3 \mathrm{ft} / \mathrm{s}\).

Step-by-Step Solution

Answer

The mass is heading downward at a velocity of 3 ft/s at \( t = \frac{5\pi}{9} \) and \( t = \frac{11\pi}{9} \).

Step 1: Set up the differential equation

The spring-mass system can be described by the second-order linear differential equation \( m\frac{d^2x}{dt^2} + kx = 0 \), where \( m \) is the mass and \( k \) is the spring constant. Given \( m = 1 \) slug and \( k = 9 \text{ lb/ft} \), the equation becomes \( \frac{d^2x}{dt^2} + 9x = 0 \).

Step 2: Determine the general solution

The general solution for the differential equation \( \frac{d^2x}{dt^2} + 9x = 0 \) is \( x(t) = C_1 \cos(3t) + C_2 \sin(3t) \). This is because the characteristic equation \( r^2 + 9 = 0 \) has roots \( r = \pm 3i \).

Step 3: Use initial conditions to find constants

The initial conditions are \( x(0) = 1 \) foot and \( \frac{dx}{dt}(0) = \sqrt{3} \text{ ft/s upward} \). Calculating the derivative, \( \frac{dx}{dt} = -3C_1 \sin(3t) + 3C_2 \cos(3t) \). Applying initial values: \( C_1 = 1 \) from \( x(0) = 1 \) and solving the velocity equation \( -3C_1 \sin(0) + 3C_2 \cos(0) = \sqrt{3} \), we find \( C_2 = \frac{\sqrt{3}}{3} \).

Step 4: Develop velocity function

The velocity function is obtained by differentiating the position function: \( \frac{dx}{dt} = -3\cos(3t) + \sqrt{3}\sin(3t) \).

Step 5: Solve for time when velocity is -3 ft/s

We need to solve \( -3\cos(3t) + \sqrt{3}\sin(3t) = -3 \), indicating a downward velocity. Rearrange and solve: \((\sin(3t) - \frac{3}{\sqrt{3}}\cos(3t) = 3)\). Dividing through by 3 gives: \( \frac{\sin(3t)}{\sqrt{3}} - \cos(3t) = -1\). Using trigonometric identities, solve for \( t \). This leads to \( t = \frac{5\pi}{9}, \frac{11\pi}{9}, \ldots \)

Key Concepts

Differential Equations

In many scientific and engineering problems, differential equations are used to describe the relationship between a function and its derivatives. Specifically, they explain how something changes over time. A second-order linear differential equation, like the one used here, typically arises in systems involving mass and elasticity, such as a spring-mass system.

Given that we have a spring-mass system, we can encounter differential equations like this:

• The general form is typically:
  \[ m\frac{d^2x}{dt^2} + kx = 0 \]
  where \( m \) represents the mass and \( k \) represents the spring constant.
• The mass, in this case, is given as \(1\) slug, and the spring constant is \(9\) lb/ft, leading us to
  \[ \frac{d^2x}{dt^2} + 9x = 0 \]

This equation models the system's mechanical oscillations, focusing on how the position changes as the mass bounces up and down, influenced by gravity and spring tension.

Initial Value Problem

An initial value problem involves finding a function which satisfies a differential equation and meets specific initial conditions. Here, these initial conditions are essential to determine the constants in the general solution of the differential equation.

Let's look at this particular problem:

• Initial displacement: The mass starts from a position 1 foot above equilibrium \( x(0) = 1 \).
• Initial velocity: The upward velocity at that moment is \( \frac{dx}{dt}(0) = \sqrt{3} \) ft/s.

These conditions dictate how the solution to the differential equation specific to this system is structured, determining the constants that appear in the specific trigonometric solution. By applying initial conditions, we unlock the behavior of the system over time.

Trigonometric Solution

To solve the differential equation for the spring-mass system, we typically find the characteristic equation, whose roots indicate the form of the solution. In this case,

• The characteristic equation:
  \[ r^2 + 9 = 0 \]
  leads to complex roots \( r = \pm 3i \), suggesting a trigonometric solution.
• This results in the general solution:
  \[ x(t) = C_1 \cos(3t) + C_2 \sin(3t) \]

This trigonometric form reflects the oscillatory motion, characteristic of systems undergoing mechanical oscillations. By using initial conditions, we find specific values for \(C_1\) and \(C_2\), making this a precise and tailored solution for the problem. The fact that we incorporate cosines and sines demonstrates how periodic oscillations are elegantly described through trigonometric functions.

Mechanical Oscillations

Mechanical oscillations refer to repetitive variations, typically in time, of some measure about a central value. For the spring-mass system, these oscillations are repetitive movements of the mass around the equilibrium point due to the forces from the spring and inertia of the mass.

Here's a breakdown of this concept:

• The position of the mass over time follows a sinusoidal path as described by the trigonometric solution:
  \[ x(t) = C_1 \cos(3t) + C_2 \sin(3t) \]
• The velocity, which shows how fast the mass is moving, changes over time as well, described by the derivative:
  \[ \frac{dx}{dt} = -3\cos(3t) + \sqrt{3}\sin(3t) \]
• Mechanical oscillations like these are influenced by factors like damping, resonance, and external forces, though in this simple system, only the spring and the mass dictate the movement.

The mass will move up and down in a regular pattern, and understanding this can simplify predicting when the mass achieves certain velocities or positions as asked in the problem. This repetitive movement is crucial to solving when the mass is moving downward at \(3 \text{ ft/s}\).