Question

A 32 lb weight stretches a spring 2 ft in equilibrium. It is attached to a dashpot with constant \(c=8\) lb-sec/ft. The weight is initially displaced 8 inches below equilibrium and released from rest. Find its displacement for \(t>0\)

Short answer

Based on the information provided in the step-by-step solution, answer the following question: **Question:** Determine the displacement function for a weight initially displaced 8 inches below equilibrium and released from rest on a spring and dashpot system with a damping constant of 8 lb-sec/ft and a spring constant of 16 lb/ft. The weight is 32 lbs. **Answer:** The displacement function for the weight movement is given by: x(t) = e^(-4t) * (-2/3 ft + (8/3 ft)*t)

Step-by-step solution

Calculate the spring constant k

Using Hooke's law, the force applied to the spring is directly proportional to the amount it is stretched. So, we can write the equation as: F = k * x Where F is the force applied (which is the weight of the object, 32 lbs), x is the stretch (2 ft), and k is the spring constant we need to find. Rearrange the equation and isolate k: k = F / x Now plug in the values of F and x from the exercise: k = 32 lbs / 2 ft Calculate the spring constant: k = 16 lb/ft

Set up the equation of motion using Newton's second law

The equation of motion for a damped spring-mass system can be written as: m * x''(t) + c * x'(t) + k * x(t) = 0 Where m is the mass causing the displacement, x(t) is the displacement function, x'(t) is the velocity function, x''(t) is the acceleration function, c is the dashpot damping constant (given as 8 lb-sec/ft), and k is the spring constant (calculated in Step 1). However, we're given the weight of the object (32 lbs), so we need to replace m with the mass. To do this, we need to use the equation: F = ma Where F is the force (weight in our case), m is the mass, and a is the acceleration due to gravity (32 ft/s^2). Rearrange this equation for m: m = F / a Now plug in the values: m = 32 lbs / 32 ft/s^2 Calculate the mass: m = 1 lb-s^2/ft Now substitute the mass into the equation of motion: 1 * x''(t) + 8 * x'(t) + 16 * x(t) = 0

Solve the differential equation for the displacement function x(t)

The given equation (1 * x''(t) + 8 * x'(t) + 16 * x(t) = 0) is a second-order linear homogeneous differential equation with constant coefficients. The general solution for this type of differential equation is given by: x(t) = e^{-rt} * (C1 * cos(wt) + C2 * sin(wt)) Where r and w depend on the values of the coefficients in the differential equation. In this case, r = -c/(2m) = -8/(2*1) = -4, so the particular solution is an exponentially decaying sinusoidal function. Also, the square root of [(-c^2/(4m^2) ) - k/m] is 0 since the function is overdamped, so the solution simplifies to: x(t) = e^{-4t} * (C1 + C2*t)

Apply initial conditions to find C1 and C2

The initial conditions are given in the exercise as: x(0) = -8 inches = -2/3 ft (Initial displacement below equilibrium) x'(0) = 0 (Initially released from rest) Apply the initial conditions to the x(t) equation: x(0) = e^{-4*0} * (C1 + C2*0) -2/3 ft = C1 Now differentiate x(t) with respect to t to get x'(t): x'(t) = -4*e^{-4*t} * (C1 + C2*t) + e^{-4t} * C2 Apply the initial condition for x'(0): x'(0) = -4*e^{-4*0} * (C1 + C2*0) + e^{-4*0} * C2 0 = -4*C1 + C2 We found C1 as -2/3 ft in the previous step, so substitute that value into the equation, and solve for C2: 0 = -4*(-2/3 ft) + C2 C2 = 8/3 ft

Write the final displacement function

With the constants C1 and C2 found, we can now write the displacement function for the weight movement for any t > 0: x(t) = e^{-4t} * (-2/3 ft + (8/3 ft)*t) This function describes the displacement of the weight released from rest and initially displaced 8 inches below equilibrium for t > 0.

Key concepts

Spring-Mass-Damper System

Understanding the mechanics of a spring-mass-damper system is essential for students studying systems that involve oscillatory motion. At its core, this system consists of three components: a spring, a mass, and a damper. The spring provides a restoring force that is proportional to the displacement, according to Hooke's law (\( F = kx \)), where F is the force exerted by the spring, k is the spring constant, and x is the displacement from the equilibrium position.

When a mass is attached to the spring, it can move freely, oscillating back and forth as the spring alternately stores and releases energy. The damper, on the other hand, dissipates energy from the system, usually in the form of heat, due to the resistance it provides to the motion. This is often modelled using a dashpot with a damping constant (\( c \)), which quantifies the amount of damping. The presence of the damper affects the character of the motion significantly, usually preventing the system from oscillating indefinitely.

The exercise provided introduces students to a fundamental application of the spring-mass-damper system in a real-world context, involving an actual weight, a spring elongation equilibrium, and a damper described by a constant. By identifying the system's parameters, students can then set up the differential equation of motion, which guides the understanding of this mechanical system's behavior over time.

Second-Order Linear Homogeneous Differential Equation

In the context of modeling the behavior of mechanical systems, second-order linear homogeneous differential equations play a crucial role. These equations take the general form \( ay''(t) + by'(t) + cy(t) = 0 \), where \( y(t) \) is the unknown function to be determined, \( y'(t) \) is the first derivative with respect to time, representing velocity, and \( y''(t) \) is the second derivative, representing acceleration. The coefficients a, b, and c are constants that describe the system’s properties.

In the solution steps for the provided exercise, we first identify the values for these constants based on the physical parameters: mass (m), damping constant (c), and spring constant (k). Then, we structure the differential equation to reflect the dynamics of the spring-mass-damper system. The solution to this particular second-order linear homogeneous differential equation reveals the motion of the system over time. By applying methods specific to this type of differential equation, such as characteristic equations or the method of undetermined coefficients, we can solve for the displacement function \( y(t) \) or \( x(t) \), which in this context are synonymous.

Damped Harmonic Motion

When delving into the behavior of physical systems that exhibit oscillations, the concept of damped harmonic motion becomes an essential topic. Damped harmonic motion describes how an object moves when influenced by a restoring force (like that from a spring) and a damping force, such as air resistance or friction from a dashpot.

In an idealized system without any damping (i.e., no energy lost), an object would oscillate indefinitely with a constant amplitude and frequency, known as simple harmonic motion. However, introducing damping into the system causes the oscillations to decay over time, eventually coming to a stop. The rate of decay depends heavily on the damping constant (\( c \) in the case of the dashpot) and the mass of the system.The solution to the exercise involves finding a specific type of solution to the differential equation that captures this damped behavior. An exponentially decaying sinusoidal function \( e^{-rt} \) is a hallmark of solutions that describe damped harmonic motion. The final displacement function \( x(t) \) showcasing the object's motion immediately indicates that the weight will not oscillate indefinitely but will come to rest at a certain point due to the damping effect. Teaching this concept helps students appreciate how mathematical models can predict physical phenomena and the effects of different forces on motion.