Question

A 64 lb object stretches a spring 4 ft in equilibrium. It is attached to a dashpot with damping constant \(c=8 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). The object is initially displaced 18 inches above equilibrium and given a downward velocity of \(4 \mathrm{ft} / \mathrm{sec}\). Find its displacement and time-varying amplitude for \(t>0 .\)

Short answer

#tag_title#7-Displacement function and time-varying amplitude#tag_content#Now that we have found the constants \(A\) and \(B\), we can write the displacement function for the damped spring system as: $$x(t) = e^{-2t}(1.5\cos(2t) + 3\sin(2t))$$ The time-varying amplitude of the oscillation is given by the magnitude of the displacement as follows: $$A(t) = \sqrt{x(t)^2} = \sqrt{e^{-4t}(2.25\cos^2(2t) + 9\cos(2t)\sin(2t) + 9\sin^2(2t))}$$ Now that we have the displacement function and time-varying amplitude, we have completed the analysis for the given spring-damper system.

Step-by-step solution

1-Determine the spring constant

To determine the spring constant, k, we use Hooke's law which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In equilibrium, the spring force is equal to the weight of the object. That is: $$kx_{eq} = W \Rightarrow k = \frac{W}{x_{eq}}$$ where \(k\) is the spring constant, \(x_{eq}\) is the equilibrium displacement, and \(W\) is the weight of the object. So, $$k=\frac{64 \ \mathrm{lb}}{4 \ \mathrm{ft}} = 16 \ \mathrm{lb/ft}$$

2-Write the equation of motion for the damped spring

Now, we can write the equation of motion for the damped spring system combining the spring force and the damping force: $$m \frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0$$ where \(m\) is the mass of the object, \(c\) is the damping constant, and \(x\) is the displacement.

3-Convert weight to mass

To convert the weight (64 lb) to mass, we'll use the conversion factor \(g \approx 32.2 \ \mathrm{ft/s^2}\): $$m = \frac{W}{g} = \frac{64 \ \mathrm{lb}}{32.2 \ \mathrm{ft/s^2}} = 2 \ \mathrm{slugs}$$

4-Involve the initial conditions

Next, we can rewrite the equation of motion using the given initial conditions: \(x(0) = 18 \ \mathrm{in} = 1.5 \ \mathrm{ft}\) and \(v(0) = 4 \ \mathrm{ft/s}\). The initial displacement is 1.5 ft and initial velocity are 4 ft/s. Thus, the equation of motion becomes: $$2\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 16x = 0 \quad , x(0)=1.5, v(0)=4$$

5-Solve the equation of motion

To solve this second-order linear differential equation with constant coefficients, we first find the characteristic equation: $$2r^2 + 8r + 16 = 0$$ This quadratic equation has two complex roots: \(r_1 = -2 + 2i\) and \(r_2 = -2 - 2i\). Therefore, the general solution for the equation of motion can be written as: $$x(t) = e^{-2t}(A\cos(2t) + B\sin(2t))$$ where \(A\) and \(B\) are constants.

6-Find constants A and B

To find the constants \(A\) and \(B\), we use the initial conditions: $$x(0) = A = 1.5$$ To find \(B\), we take the first derivative of the displacement function: $$v(t) = \frac{dx}{dt} = -2e^{-2t}(A\cos(2t) + B\sin(2t)) + e^{-2t}(-2A\sin(2t) + 2B\cos(2t))$$ Using the initial condition \(v(0) = 4 \ \mathrm{ft/s}\), we have: $$4 = -2A + 2B \Rightarrow B = 3\text{ }.$$

Key concepts

Spring Constant

A spring constant, denoted as \( k \), is a measure of a spring's stiffness. It's crucial in harmonic motion as it determines how much force is needed to stretch or compress a spring by a certain distance. Hooke's Law provides the foundation for calculating the spring constant: \( F = kx \), where \( F \) is the force applied, and \( x \) is the displacement from the equilibrium position. This means that the spring force is directly proportional to its displacement.
In the given problem, the object weighs 64 lb and stretches the spring by 4 ft in equilibrium, allowing us to calculate the spring constant as follows:

• Weight \( W = 64 \) lb
• Displacement \( x_{eq} = 4 \) ft

Using the formula \( k = \frac{W}{x_{eq}} \), we derive that \( k = 16 \text{ lb/ft} \).
Understanding the spring constant helps us predict how the spring will behave under different forces, a fundamental aspect of analyzing harmonic systems.

Differential Equations

Differential equations are essential in modeling real-world phenomena such as damped harmonic motion. They contain derivatives which represent rates of change, making them effective for describing dynamic systems. In the context of harmonic motion, we deal with second-order linear differential equations with constant coefficients.
The general form of the equation for a damped spring system is:
\[m \frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0\]
Each part of the equation represents different forces:

• \( m \frac{d^2x}{dt^2} \): The mass \(m\) and the acceleration \( \frac{d^2x}{dt^2} \), together representing the inertial force.
• \( c\frac{dx}{dt} \): The damping force with damping constant \(c\), which opposes the motion.
• \( kx \): The spring force due to displacement, based on the spring constant \(k\).

Solving differential equations involves finding functions that satisfy the equation under given conditions. In our case, the relevant differential equation provides the displacement \(x(t)\) of the object over time.

Initial Conditions

Initial conditions are the specific parameters set at the beginning of the time period we're analyzing. They're vital in determining how the system evolves over time. In the scenario of damped harmonic motion, initial conditions often include the initial displacement and velocity of the object.
For this exercise, the initial conditions provided are:

• Initial displacement, \( x(0) = 1.5 \text{ ft} \) (since 18 inches equals 1.5 feet)
• Initial velocity, \( v(0) = 4 \text{ ft/s} \)

These conditions describe the system's state at \( t = 0 \).
Incorporating these initial conditions into the solution of the differential equations allows us to find specific values for constants in the general solution. For instance, by applying \( x(0) \) and \( v(0) \) to the solution \( x(t) = e^{-2t}(A\cos(2t) + B\sin(2t)) \), we determine \( A = 1.5 \) and \( B = 3 \).
Having initial conditions ensures that our mathematical model replicates the actual physical scenario accurately.