Question

An 8-lb weight is attached to the lower end of a coil spring suspended from the ceiling and comes to rest in its equilibrium position, thereby stretching the spring \(0.4 \mathrm{ft}\). The weight is then pulled down 6 in. below its equilibrium position and released at \(t=0\). The resistance of the medium in pounds is numerically equal to \(2 x^{\prime}\), where \(x^{\prime}\) is the instantaneous velocity in feet per second. (a) Set up the differential equation for the motion and list the initial conditions. (b) Solve the initial-value problem set up in part (a) to determine the displacement of the weight as a function of the time. (c) Express the solution found in step (b) in the alternative form (5.32) of the text. (d) What is the quasi period of the motion? (e) Graph the displacement as a function of the time.

Short answer

The displacement function for the given problem is: \[x(t) = 0.502 e^{-\frac{1}{8}t} \cos(\frac{15}{4}t - 88.18) + \frac{g}{20}\] The quasi period of the motion is approximately \(1.677 \ \mathrm{s}\).

Step-by-step solution

Set up the differential equation

According to Newton's second law, the net force on the weight is proportional to the mass times the acceleration. The forces acting on the weight include the spring force, gravitational force, and damping force due to the medium. The spring force can be modeled as \(-kx\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position. The damping force due to the medium is given as \(2 x'\), where \(x'\) is the velocity of the weight. The net force acting on the weight is: \[F_{net} = -kx - mg - 2x'\] Since \(F_{net} = ma\), and \(a = x''\) (the acceleration is the second derivative of the displacement), we can write the differential equation as: \[m x'' = -kx -mg - 2 x'\] Initially, the weight is in equilibrium when stretched by \(0.4 \mathrm{ft}\), so \(k(0.4) = mg\), or \(k = 20 \times 8 = 160 \mathrm{lb/ft}\). Now we can substitute this value of k in our differential equation: \[m x'' = -160x -mg - 2 x'\] Since m = 8 lb, we finally have the differential equation: \[x'' = -20x -g - \frac{1}{4} x'\]

List the initial conditions

The initial conditions for the problem are: The weight is pulled down \(6 \ \mathrm{in.} = 0.5\mathrm{ft}\) below the equilibrium position and released with \(t=0\). Thus, \[x(0) = 0.5 \ \mathrm{ft}\] \[x'(0) = 0 \ \mathrm{ft/s}\]

Solve the initial value problem

To solve the initial value problem, we have the linear second-order differential equation: \[x'' = -20x -g - \frac{1}{4} x'\] We can use Laplace transform to solve this equation. Assuming the Laplace transform of \(x(t)\) is \(X(s)\), we get: \[s^2X(s) - x(0)s - x'(0) = -20X(s) -g - \frac{1}{4} (sX(s) - x(0))\] Plugging in the initial conditions, we have: \[s^2X(s) - 0.5s = -20X(s) -g - \frac{1}{4} sX(s) + \frac{1}{8}\] Now, solve for \(X(s)\): \[X(s) = \frac{0.5s +g + \frac{1}{8}}{s^2 + \frac{1}{4}s + 20}\] Finally, apply inverse Laplace transform to find \(x(t)\): \[x(t) = L^{-1}\{X(s)\} = L^{-1}\left\{\frac{0.5s +g + \frac{1}{8}}{s^2 + \frac{1}{4}s + 20}\right\}\] By partial fraction decomposition and applying inverse Laplace transform, we can find the displacement function \(x(t)\): \[x(t) = C_1 e^{-\frac{1}{8}t} \cos(\frac{15}{4}t) + C_2 e^{-\frac{1}{8}t} \sin(\frac{15}{4}t) + \frac{g}{20}\] Now, using the initial conditions, we can solve for constants \(C_1\) and \(C_2\): \[x(0) = C_1 = 0.5 \ \mathrm{ft}\] \[x'(0) = -\frac{1}{8} C_1 + \frac{15}{4} C_2 = 0\] Solving for the above system of equations, we get \(C_1 = 0.5 \ \mathrm{ft}\) and \(C_2 = \frac{1}{30} \ \mathrm{ft/s}\). Therefore, the displacement function is: \[x(t) = 0.5 e^{-\frac{1}{8}t} \cos(\frac{15}{4}t) + \frac{1}{30} e^{-\frac{1}{8}t} \sin(\frac{15}{4}t) + \frac{g}{20}\]

Find the alternative representation and the quasi period

We can write the displacement function in the alternative form: \[x(t) = A e^{-\frac{1}{8}t} \cos(\frac{15}{4}t - \phi) + \frac{g}{20}\] Here, the amplitude \(A\) can be found using \(A^2 = C_1^2 + C_2^2\), and \(\phi\) using \(\tan \phi = \frac{C_1}{C_2}\). The amplitude is given by: \[A = \sqrt{0.5^2 + \frac{1}{30^2}} \approx 0.502 \ \mathrm{ft}\] The phase angle is given by: \[\tan \phi = \frac{0.5}{\frac{1}{30}} \implies \phi \approx 88.18^\circ\] And the quasi period is given by: \[T = \frac{2\pi}{\frac{15}{4}} \approx 1.677 \ \mathrm{s}\]

Graph the displacement as a function of the time

To graph the displacement function \(x(t)\), we can use a graphing software or calculator to plot: \[x(t) = 0.502 e^{-\frac{1}{8}t} \cos(\frac{15}{4}t - 88.18) + \frac{g}{20}\] The displacement should show an oscillatory motion with a decreasing amplitude due to the damping effect of the medium, and the equilibrium position at \(\frac{g}{20}\).

Key concepts

Initial Value Problem

An initial value problem in the context of differential equations involves finding a particular solution to a differential equation that also satisfies specific conditions set at the initial point. In this exercise, we consider the motion of a weight attached to a spring. The problem is to determine the equation of motion, which is a second-order differential equation, and solve it using the given initial conditions. These conditions include the initial displacement of the weight from its equilibrium position and its initial velocity.

In the problem, the initial conditions are: the weight is displaced 6 inches (or 0.5 feet) below the equilibrium, and it is released at rest, meaning the initial velocity is zero. The task requires setting up both the equation and solving it to find how the weight moves over time.

Laplace Transform

The Laplace transform is a powerful mathematical technique used to simplify the process of solving differential equations, especially in engineering and physics applications. It converts differential equations, which are often complex and difficult to solve directly, into simpler algebraic forms in the Laplace domain.

In this exercise, once the differential equation for the weight's motion is established, the Laplace transform is applied. By transforming the original equation into the Laplace form, solving for the Laplace transform of the displacement becomes an algebraic manipulation problem rather than a calculus one.

• Apply the Laplace transform to both sides of the differential equation.
• Use initial conditions provided in the problem to solve for constants.
• Manipulate the resulting algebraic equation to isolate the transformed variable.
• Use the inverse Laplace transform to revert back to the time domain and find the function describing displacement over time.

Damping Force

Damping force represents the resistance opposing the motion of an oscillating system. In this exercise, it is expressed as being proportional to the velocity of the weight as it moves through the air. The equation given for the damping force is numerical and equal to twice the velocity of the weight at any instant, denoted as \(2x'\), with \(x'\) being the velocity.

• The damping force acts in a direction opposite to the velocity, slowing down the motion of the system.
• Incorporating damping force in the differential equation models realistic physical systems where energy loss is inevitable due to resistive forces.
• The damping causes the amplitude of the oscillation to decrease over time, which leads to the eventual settling of the system back to its equilibrium position.

Equilibrium Position

Equilibrium position in this context refers to the position where the forces acting on the weight are balanced such that the net force is zero. At equilibrium, the spring force equals the gravitational force, and the system is at rest unless acted upon by an external displacement.

In this exercise, the weight stretches the spring by 0.4 feet to reach the equilibrium position initially. Any displacement from this point initiates oscillations of the weight.

• The spring force always acts towards restoring the weight to the equilibrium position.
• Equilibrium is affected by the spring constant and the weight of the object, as demonstrated by the relation \(k \times 0.4 = mg\).
• Understanding equilibrium helps in setting up the differential equation that governs the motion of the system.

By grasping the concept of equilibrium, one can better understand how external forces and initial conditions influence the overall motion dynamics of the system.