Question

An 8 lb weight stretches a spring 2 inches. It is attached to a dashpot with damping constant \(c=4 \mathrm{lb}-\mathrm{sec} / \mathrm{ft}\). The weight is initially displaced 3 inches above equilibrium and given a downward velocity of \(4 \mathrm{ft} / \mathrm{sec} .\) Find its displacement for \(t>0\)

Short answer

The displacement function of the 8 lb weight attached to the spring and dashpot system at any time \(t>0\) is: $$x(t) = \frac{3}{4} e^{-8t} - \frac{1}{2} e^{-2t}$$

Step-by-step solution

Calculate the spring constant

To find the spring constant, we can use Hooke's law, which states that the force exerted by a spring is proportional to the displacement from its equilibrium position. In this case, we are given that an 8 lb weight stretches the spring by 2 inches (1/6 ft). The equation for Hooke's law is: $$F = kx$$ Where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement. Rearranging this equation, we can find the spring constant: $$k = \frac{F}{x} = \frac{8}{1/6}=48\, \mathrm{lb} / \mathrm{ft}$$

Write the equation of motion

In order to find the displacement, we need to set up and solve the equation of motion of the damped system. The equation of motion for a damped system is given by the following ordinary differential equation: $$m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx=0$$ We are given that the weight is 8 lb and the damping constant is 4 lb-sec/ft. Using the conversion factor 1 lb = 32.174 ft/s^2, we can find the mass: $$m = \frac{8}{32.174} = 0.2486 \, \mathrm{slugs}$$ Now, we can substitute the values of \(m\), \(c\), and \(k\): $$0.2486 \frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 48x=0$$

Solve the equation of motion

To solve the equation of motion, we can assume that the displacement \(x(t)\) has the form \(x(t)=e^{rt}\). Differentiating \(x(t)\) twice: $$\frac{dx}{dt}=re^{rt}\quad\text{and}\quad\frac{d^2x}{dt^2}=r^2e^{rt}$$ Substitute these expressions into the differential equation: $$0.2486(r^2e^{rt})+4(re^{rt})+48(e^{rt})=0$$ Divide through by \(e^{rt}\): $$0.2486r^2+4r+48=0$$ Using quadratic formula to solve for \(r\): $$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{(-4)^2 - 4(0.2486)(48)}}{2(0.2486)}$$ Solving for \(r\), we get two roots, \(r_1=-8\) and \(r_2=-2\). Thus, the general solution is: $$x(t)=C_1 e^{-8t}+C_2 e^{-2t}$$

Apply initial conditions

The initial conditions are given as: 1. Initial displacement, \(x(0) = 3\) inches (1/4 ft) 2. Initial velocity, \(\frac{dx(0)}{dt} = 4 \, \mathrm{ft} / \mathrm{sec}\) Applying the first initial condition: $$x(0) = C_1 e^{-8(0)} + C_2 e^{-2(0)} = C_1 + C_2 = \frac{1}{4}$$ Applying the second initial condition: $$\frac{dx(t)}{dt}=-8C_1 e^{-8t} - 2C_2 e^{-2t}$$ $$\frac{dx(0)}{dt} = -8C_1 - 2C_2 = 4$$ With the system of equations: $$\begin{cases} C_1 + C_2 = \frac{1}{4}\\ -8C_1 - 2C_2 = 4 \end{cases}$$ Solving for the constants \(C_1\) and \(C_2\), we get: $$C_1 = \frac{3}{4}\quad\text{and}\quad C_2 = -\frac{1}{2}$$

Write the final displacement function

Substituting the obtained values for \(C_1\) and \(C_2\) into \(x(t)\) yields the final displacement function: $$x(t)=\frac{3}{4} e^{-8t} - \frac{1}{2} e^{-2t}$$ This is the displacement function of the 8 lb weight attached to the spring and dashpot system at any time \(t>0\).

Key concepts

Damped Harmonic Oscillator

A damped harmonic oscillator is a system consisting of a mass, a spring, and a damping element, such as a dashpot. This setup allows the system to oscillate, but with the damping effect gradually reducing the amplitude of vibrations over time. The damping force is usually proportional to the velocity of the mass, opposing the motion and transforming energy into heat or other forms. The system can be described with parameters like mass (m), spring constant (k), and damping constant (c). These elements work together to create an oscillatory motion that gradually fades, demonstrating real-world scenarios where friction and resistance are present.

Initial Conditions

Initial conditions in differential equations specify the starting state of a system. For oscillating systems, initial conditions often include the initial displacement and initial velocity. These values are crucial as they help determine the specific solution from a family of possible solutions.
In the context of the exercise, the initial displacement tells us how far the weight was initially moved from equilibrium, while the initial velocity indicates the speed and direction of initial movement. Together, these conditions inform us about the system's behavior at the onset and allow us to tailor our general solution to match the real-world setup.

Ordinary Differential Equation (ODE)

An ordinary differential equation (ODE) involves functions of a single independent variable and their derivatives. In the context of the damped harmonic oscillator, the ODE captures the interactions between the mass, spring, and damping elements through derivatives that reflect acceleration and velocity changes.
This ODE generally takes the form: \[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0 \]Here, the terms represent acceleration, damping, and spring force, respectively. Solving this ODE involves finding a function that describes the displacement over time, reflecting how the system evolves.

Equation of Motion

The equation of motion is a mathematical expression that describes the dynamic behavior of a system. For a damped harmonic oscillator, the equation of motion is formed by balancing forces acting on the system, including inertia, damping, and restoring forces from the spring.

• Inertial force is linked to acceleration through mass.
• Damping force depends on velocity and opposes motion.
• Restoring force springs from Hooke’s law, related to displacement.

These forces interrelate to form the basis for the ODE, encapsulating all mechanical dynamics at play. Solving the equation of motion gives the displacement as a function of time, offering insights into how oscillations decay due to damping.