Problem 21 - Advanced Engineering Mathematics

Question

A mass weighing 4 pounds is attached to a spring whose constant is \(2 \mathrm{lb} / \mathrm{ft}\). The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 foot above the equilibrium position with a downward velocity of \(8 \mathrm{ft} / \mathrm{s}\). Determine the time at which the mass passes through the equilibrium position. 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 mass passes equilibrium at \( t = \frac{1}{4} \) s and reaches extreme displacement at \( t = \frac{1}{2} \) s, with position \( x = 0 \).

Step 1: Identify the parameters and form the differential equation

The mass is 4 pounds and the spring constant is 2 lb/ft. The damping coefficient (numerically equal to velocity) is 1 since it equals the instantaneous velocity. Therefore, the mass of the object in slugs is calculated as \( m = \frac{4}{32} = \frac{1}{8} \) slugs (weight divided by gravity, 32 ft/s²). The system is underdamped, and the differential equation of motion can be written as \( \frac{1}{8}x''(t) + x'(t) + 2x(t) = 0 \) for the damped harmonic oscillator.

Step 2: Convert to a standard form

Multiply the entire differential equation by 8 to obtain a more manageable form: \( x''(t) + 8x'(t) + 16x(t) = 0 \). This is a second-order linear homogenous differential equation.

Step 3: Solve the characteristic equation

The characteristic equation associated with the differential equation is \( r^2 + 8r + 16 = 0 \). Solving this quadratic equation, we find the roots \( r = -4 \) (a double root), indicating a critically damped system.

Step 4: Write the general solution of the differential equation

For a critically damped system with repeated roots, the solution is given by \( x(t) = (c_1 + c_2 t) e^{-4t} \), where \( c_1 \) and \( c_2 \) are constants to be determined from the initial conditions.

Step 5: Apply initial conditions to find constants

Given the initial conditions: \( x(0) = 1 \) (initial displacement) and \( x'(0) = -8 \) (initial velocity downward since velocity is taken as negative for downward direction). Use these conditions to set up equations: \( 1 = (c_1)(1) \) yields \( c_1 = 1 \). Differentiating the solution to get velocity: \( x'(t) = -4(c_1 + c_2 t) e^{-4t} + c_2 e^{-4t} \). At \( t=0 \), \( x'(0) = -4c_1 + c_2 = -8 \), substituting \( c_1 = 1 \) gives \( -4 + c_2 = -8 \), so \( c_2 = -4 \).

Step 6: Write the specific solution with constants

Substitute \( c_1 = 1 \) and \( c_2 = -4 \) into the solution: \( x(t) = (1 - 4t)e^{-4t} \).

Step 7: Determine when the mass passes through the equilibrium position

The mass passes through the equilibrium position when \( x(t) = 0 \). Set \( (1 - 4t)e^{-4t} = 0 \). The exponential function is never zero, so \( 1 - 4t = 0 \). Solving for \( t \), we find \( t = \frac{1}{4} \) seconds.

Step 8: Determine the extreme displacement and its time

The extreme displacement is found by setting the derivative of \( x(t) \) equal to zero: solve \( x'(t) = 0 \). From earlier, \( x'(t) = (-4(1-4t) - 4)e^{-4t} \), setting \( -4(1-4t) - 4 = 0 \), we find \( t = \frac{1}{2} \) seconds. Use the specific solution \( x(t) = (1-4t)e^{-4t} \) to find the displacement \( x\left(\frac{1}{2}\right) = (1 - 4 \cdot \frac{1}{2})e^{-2} = 0 \).

Key Concepts

Damped Harmonic Oscillator

The damped harmonic oscillator is a system in which there is a restoring force proportional to displacement but also a damping force opposing the motion, proportional to velocity. This is common in scenarios with springs and friction. In our exercise, the equation

• \[ m x''(t) + c x'(t) + k x(t) = 0 \]

represents it, where:

• \( m \): mass
• \( c \): damping coefficient
• \( k \): spring constant

Differential equations like this describe how the motion evolves over time, often leading to exponentially decaying oscillations if there’s damping. Understanding this can help predict how systems like car suspensions or building foundations behave, ensuring safety and comfort.

Initial Conditions

Initial conditions provide the specific starting point for solving differential equations. In dynamic systems like a damped harmonic oscillator, initial displacement and velocity determine how the motion will unfold. They enable us to find the unique solution out of infinitely many possibilities that align with these starting states.

For example, given:

• Initial displacement: 1 foot above equilibrium
• Initial velocity: 8 ft/s downward

these are plugged into the general solution to find precise constants that fully describe the motion specific to those starts. Recognizing the role of initial conditions is crucial as they tailor the system’s behavior to match real-life situations, like launching a rocket or bringing a car to a stop.

Equilibrium Position

The equilibrium position in oscillatory systems is the point where the net forces on the mass are zero. For a mass-spring system, this is where the spring force balances the weight and no motion would occur if left undisturbed.

It's an important concept because solutions often relate to how deviations from equilibrium change over time. In our calculation, solving for when the mass returns to equilibrium gives:

• The time when oscillatory up and down motion stabilizes at \( t = \frac{1}{4} \) seconds.

Understanding equilibrium helps predict how systems respond to perturbations—in engineering, keeping structures stable in their equilibrium can prevent disasters.

Extreme Displacement

Extreme displacement refers to the point at which the system reaches its maximum or minimum position from the equilibrium. In physical systems, it marks the peak of oscillation before returning due to restoring forces.

To find these points, we use calculus to set the derivative of position to zero, determining when the velocity (rate of change of position) momentarily stops before reversing. In our example, this led to:

• Extreme displacement occurs at \( t = \frac{1}{2} \) seconds.

Knowing these maxima is essential in design and testing of products subject to vibrations, helping ensure they can withstand or efficiently respond to those forces without failure.