Problem 97 - Calculus 3

Question

A 64-lb weight is attached to a vertical spring with a spring constant of 4.625 lb/ft. The resistance in the spring-mass system is equal to the instantaneous velocity. The weight is set in motion from a position 1 ft below its equilibrium position with an upward velocity of 2 ft/sec. Is the mass above or below the equation position at the end of \(\pi\) sec? By what distance?

Step-by-Step Solution

Answer

The mass is above the equilibrium position by approximately 0.59 ft at \(t = \pi\) seconds.

Step 1: Understand the Problem

We need to determine the position of the weight after a time of \(\pi\) seconds, given the initial conditions of a spring-mass system. The weight starts 1 foot below equilibrium with an upward velocity of 2 ft/sec and a damping force equal to the velocity.

Step 2: Write the Governing Differential Equation

The system is described by the second-order linear differential equation: \[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \]where \(m\) is the mass, \(c\) is the damping coefficient, \(k\) is the spring constant, and \(x(t)\) is the displacement. For this problem, \(m = \frac{64}{32.2} \approx 1.988 \text{ slugs}\), \(c = 1\), and \(k = 4.625\).

Step 3: Solve the Characteristic Equation

The characteristic equation is derived from the differential equation: \[ m\lambda^2 + c\lambda + k = 0 \]Substituting the values gives: \[ 1.988\lambda^2 + \lambda + 4.625 = 0 \]We solve for the roots using the quadratic formula: \[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1.988\), \(b = 1\), and \(c = 4.625\).

Step 4: Calculate the Roots

Calculating the discriminant: \[ 1^2 - 4 \times 1.988 \times 4.625 = \Delta \]\(\Delta < 0\), indicating complex roots: \[ \lambda = \alpha \pm i \beta \] Find \(\alpha = -\frac{1}{2 \times 1.988}\) and \(\beta = \sqrt{\lvert \Delta\rvert}\).

Step 5: Write the General Solution

The general solution for displacement is: \[ x(t) = e^{-\alpha t}[C_1 \cos(\beta t) + C_2 \sin(\beta t)] \] Substitute \(\alpha\) and \(\beta\) from the roots found in Step 4.

Step 6: Apply Initial Conditions

The initial conditions are \(x(0) = -1\) and \(\frac{dx}{dt}(0) = 2\). Use these to solve for \(C_1\) and \(C_2\) in the general solution, yielding a particular solution.

Step 7: Evaluate the Position at \(t = \pi\)

Substitute \(t = \pi\) into the particular solution for \(x(t)\) to find the displacement. This requires calculating the cosine and sine of appropriate terms and applying the exponential decay.

Step 8: Interpret the Result

Determine if the displacement at \(t = \pi\) is positive or negative to conclude whether the mass is above or below the equilibrium position. Compute the magnitude to find the exact distance.

Key Concepts

Differential Equation

In a spring-mass system, we often encounter differential equations. These equations allow us to model the dynamics of the system. For such systems, the second-order linear differential equation is typically used. This equation can be written as follows:

• The mass term: \(m \frac{d^2x}{dt^2}\) represents the inertia of the mass as it moves.
• The damping term: \(c \frac{dx}{dt}\) accounts for the resistance that is proportional to the velocity.
• The spring term: \(kx\) is the force exerted by the spring, which is proportional to the displacement from the equilibrium position.

Using these components, the equation is structured as \( m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \). Here, \(m\) is the mass of the weight in the system, \(c\) is the damping coefficient, and \(k\) is the spring constant. This equation generally describes the forces in the system being in equilibrium, meaning their sum is zero.

Damping Force

The damping force is a crucial part of many spring-mass systems, especially when examining real-world physical phenomena. Damping refers to the resistance that counteracts the motion of a mass connected to a spring. In essence, it reduces the amplitude of oscillations over time.There are a few key points to understand about damping:

• Damping is a force that is dependent on the velocity; as velocity increases, so does the damping force.
• In the context of the exercise, the damping force is equal to the instantaneous velocity. This feature simplifies the damping term in the differential equation to \( c = 1 \).
• Such systems are called underdamped, particularly when the damping force is not strong enough to bring the system to rest quickly.

The damping force plays a vital role in determining the system's behavior over time—whether it will continue oscillating, stop quickly, or somewhere in between.

Characteristic Equation

The characteristic equation arises from the differential equation governing the spring-mass system. It provides the key to solving the differential equation by allowing us to determine the system's natural frequencies.The given differential equation:\[ m\lambda^2 + c\lambda + k = 0 \]is a quadratic equation in terms of \(\lambda\). The roots of this quadratic equation indicate the type of response the system will have.

• For our school exercise, solving this equation involves substituting the values of mass \(m = 1.988\), damping coefficient \(c = 1\), and spring constant \(k = 4.625\).
• Using the quadratic formula \( \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the discriminant \(\Delta = b^2 - 4ac\) reveals the nature of the roots.
• A negative discriminant indicates complex roots, suggesting an oscillatory motion typical of underdamped systems.

By calculating these roots, we can proceed to find the solution for displacement over time.

Initial Conditions

Initial conditions are the specific values of the displacement and velocity of the system at the beginning of observation, i.e., \(t = 0\). These conditions are crucial to finding a particular solution to the differential equation, as they help to determine the exact nature of the motion in a time-dependent manner.In this problem:

• The initial displacement \(x(0) = -1\) ft indicates that the weight is initially positioned 1 ft below its equilibrium state.
• The initial velocity \(\frac{dx}{dt}(0) = 2\) ft/sec conveys an upward motion, serving as a starting impulse for the system.

Using these initial conditions, we substitute them into the general solution to find constants \(C_1\) and \(C_2\). These constants define the particular solution to the differential equation, reflecting the unique impact the initial conditions have on the system's movement. Ultimately, evaluating this particular solution at any given time \(t\) allows us to determine the position of the mass, thus fully describing the dynamics of the system.