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Magazine Chapter 5: Problem 20
An object weighing \(8 \mathrm{lb}\) stretches a spring 6 in beyond its natural length. If the resistive force is \(F_{R}=4 d x / d t\), find the displacement of the object if it is set into motion from its equilibrium position with an initial velocity of \(1 \mathrm{ft} / \mathrm{s}\) in the downward direction.
Short Answer
Expert verified
The displacement is described by the function \( x(t) = -te^{-8t} \).
Step by step solution
01
Convert Units
First, convert the weight of the object into mass. Since 1 lb is equal to the force of gravity acting on 1 lb of mass, the weight (force) is 8 lb. We can find the mass by dividing the weight by the acceleration due to gravity, which is 32 ft/s虏. So, the mass is \[ m = \frac{8 \text{ lb}}{32 \text{ ft/s}^2} = 0.25 \text{ slugs} \]. Note that 6 inches must also be converted into feet: \[ 6 \text{ inches} = 0.5 \text{ feet} \].
02
Determine the Spring Constant
Use Hooke's Law to find the spring constant \( k \). Hooke's Law states that \[ F = kx \]. Given that a force of 8 lb stretches the spring by 0.5 ft, we can solve for \( k \): \[ 8 \text{ lb} = k \times 0.5 \text{ ft} \implies k = 16 \text{ lb/ft} \].
03
Write the Differential Equation
The system can be modeled by the second-order linear differential equation \[ m \frac{d^2x}{dt^2} + F_R + kx = 0 \]. Given that the resistive force is \( F_R = 4 \frac{dx}{dt} \), substitute \( m = 0.25 \text{ slugs} \) and \( k = 16 \text{ lb/ft} \) into the equation: \[ 0.25 \frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 16x = 0 \].
04
Simplify the Differential Equation
To simplify, multiply through by 4 to eliminate the fraction: \[ \frac{d^2x}{dt^2} + 16 \frac{dx}{dt} + 64x = 0 \]. This is a standard form of a second-order differential equation.
05
Solve the Characteristic Equation
The characteristic equation corresponding to the differential equation is: \[ r^2 + 16r + 64 = 0 \].This can be solved using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1 \), \( b = 16 \), and \( c = 64 \). Calculating the roots, \[ r = \frac{-16 \pm \sqrt{16^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} = \frac{-16 \pm 0}{2} = -8 \]. Thus, the equation has a repeated root of \( r = -8 \).
06
Write the General Solution
For a differential equation with repeated roots, the general solution takes the form: \[ x(t) = (C_1 + C_2t)e^{-8t} \].
07
Apply Initial Conditions
Use the initial conditions to solve for \( C_1 \) and \( C_2 \). Initially, the displacement is zero, \( x(0) = 0 \), and the initial velocity is 1 ft/s downward, \( x'(0) = -1 \):1. For \( t = 0 \): \[ x(0) = (C_1 + C_2 \cdot 0)e^{-8 \cdot 0} = C_1 = 0 \].2. For velocity, differentiate and substitute \( t = 0 \):\[ x'(t) = (C_2 - 8(C_1 + C_2t))e^{-8t} \implies x'(0) = C_2 - 8C_1 = -1 \]. Since \( C_1 = 0 \), this simplifies to \( C_2 = -1 \).
08
Write Final Solution
Substitute the values of \( C_1 \) and \( C_2 \) into the general solution:\[ x(t) = (-1t)e^{-8t} \].Thus, the displacement as a function of time is:\[ x(t) = -te^{-8t} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
spring constant calculation
To solve problems involving springs, we often start by calculating the spring constant, denoted as \( k \). This constant represents the stiffness of the spring. According to Hooke's Law, the force exerted by a spring is directly proportional to the displacement from its natural length. The formula is:
\[ F = kx \].
In this exercise, we are given that an 8 lb weight stretches a spring by 6 inches (or 0.5 feet). Rearranging Hooke's Law to solve for \( k \), we have:
\[ k = \frac{F}{x} = \frac{8 \text{ lb}}{0.5 \text{ ft}} = 16 \text{ lb/ft} \].
Thus, the spring constant is 16 lb/ft. Calculating the spring constant is crucial because it allows us to set up the differential equation that describes the system's motion.
\[ F = kx \].
In this exercise, we are given that an 8 lb weight stretches a spring by 6 inches (or 0.5 feet). Rearranging Hooke's Law to solve for \( k \), we have:
\[ k = \frac{F}{x} = \frac{8 \text{ lb}}{0.5 \text{ ft}} = 16 \text{ lb/ft} \].
Thus, the spring constant is 16 lb/ft. Calculating the spring constant is crucial because it allows us to set up the differential equation that describes the system's motion.
second-order linear differential equation
With the spring constant calculated, our next step is to model the motion of the object attached to the spring using a second-order linear differential equation. This type of equation typically describes systems involving acceleration, velocity, and displacement. The general form of the equation for a damped harmonic oscillator is:
\[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \].
Where:
In our exercise, \(m = 0.25\) slugs, \(c = 4\), and \(k = 16\). Substituting these values, we get:
\[ 0.25 \frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 16x = 0 \].
To eliminate the fraction, multiply the entire equation by 4: \[ \frac{d^2x}{dt^2} + 16 \frac{dx}{dt} + 64x = 0 \].
This is a second-order linear differential equation with constant coefficients. Solving this equation provides insight into the motion of the spring-mass system.
\[ m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \].
Where:
- \( m \) is the mass
- \( c \) is the damping coefficient
- \( k \) is the spring constant
- \( x \) is the displacement
In our exercise, \(m = 0.25\) slugs, \(c = 4\), and \(k = 16\). Substituting these values, we get:
\[ 0.25 \frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 16x = 0 \].
To eliminate the fraction, multiply the entire equation by 4: \[ \frac{d^2x}{dt^2} + 16 \frac{dx}{dt} + 64x = 0 \].
This is a second-order linear differential equation with constant coefficients. Solving this equation provides insight into the motion of the spring-mass system.
initial conditions in differential equations
Initial conditions are vital in solving differential equations because they allow us to find specific solutions that describe real-world scenarios. For our spring problem, we need to determine the constants in the general solution. The initial conditions provided are:
The general solution for the differential equation with a repeated root \( r = -8 \) is: \[ x(t) = (C_1 + C_2t)e^{-8t} \].
Using the initial conditions, we set:
Thus, \( C_1 = 0 \) and \( C_2 = -1 \). The final solution is: \[ x(t) = -te^{-8t} \].
Initial conditions are essential steps to ensure the solutions accurately represent the physical system being modeled.
- The initial displacement, \( x(0) = 0 \)
- The initial velocity is 1 ft/s downward, \( x'(0) = -1 \)
The general solution for the differential equation with a repeated root \( r = -8 \) is: \[ x(t) = (C_1 + C_2t)e^{-8t} \].
Using the initial conditions, we set:
- \( x(0) = C_1 = 0 \)
- \( x'(0) = C_2 - 8C_1 = C_2 = -1 \)
Thus, \( C_1 = 0 \) and \( C_2 = -1 \). The final solution is: \[ x(t) = -te^{-8t} \].
Initial conditions are essential steps to ensure the solutions accurately represent the physical system being modeled.
