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Magazine Chapter 10: Problem 2
A spring is such that a 2 -lb weight stretches it \(\frac{1}{2}\) ft. An impressed force \(\frac{1}{4} \sin 8 t\) and a damping force of magnitude \(|v|\) are both acting on the spring. The weight starts \(\frac{1}{4}\) ft below the equilibrium point with an imparted upward velocity of \(3 \mathrm{ft} / \mathrm{sec} .\) Find a formula for the position of the weight at time \(t\).
Short Answer
Expert verified
The position of the weight at time \( t \) is \( y(t) = \left(-\frac{19}{60} + C_2 t \right) e^{-8t} - \frac{1}{60} \sin 8t \).
Step by step solution
01
Calculate the Spring Constant
According to Hooke's Law, the force exerted on a spring is proportional to the displacement. Here, 2 lbs stretches the spring by \( \frac{1}{2} \) ft. Thus, the spring constant \( k \) is \( k = \frac{2}{\frac{1}{2}} = 4 \).
02
Establish the Differential Equation
The spring-mass-damping-impressed force system can be described by the differential equation: \[ m \cdot \frac{d^2y}{dt^2} + c \cdot \frac{dy}{dt} + ky = F(t) \]where \( F(t) = \frac{1}{4} \sin 8t \), \( c = 1 \) (from the damping force \(|v|\)), and \( k = 4 \). The weight is 2 lbs, which converts to mass \( m = \frac{2}{32} = \frac{1}{16} \) slugs, leading to:\[ \frac{1}{16} \frac{d^2y}{dt^2} + \frac{dy}{dt} + 4y = \frac{1}{4} \sin 8t \]
03
Convert to a Simple Form
Multiply the entire differential equation by 16 to remove the fraction:\[ \frac{d^2y}{dt^2} + 16 \frac{dy}{dt} + 64y = 4 \sin 8t \]
04
Determine Initial Conditions
Given initial position \( y(0) = -\frac{1}{4} \) ft (below the equilibrium) and initial velocity \( v(0) = 3 \) ft/s (upwards), we have:\[ y(0) = -\frac{1}{4}, \quad y'(0) = 3 \]
05
Solve the Homogeneous Equation
The characteristic equation is:\[ r^2 + 16r + 64 = 0 \]which factors as:\[ (r + 8)^2 = 0 \]So, \( r = -8 \) has multiplicity 2. The homogeneous solution is then:\[ y_h(t) = (C_1 + C_2 t) e^{-8t} \]
06
Solve the Particular Solution
For the particular solution of \( 4 \sin 8t \), assume a form:\[ y_p(t) = A \sin 8t + B \cos 8t \]Substitute into the differential equation to find \( A \) and \( B \). After calculations, find:\[ A = -\frac{1}{60}, \quad B = 0 \]So:\[ y_p(t) = -\frac{1}{60} \sin 8t \]
07
Combine Solutions and Apply Initial Conditions
The general solution is:\[ y(t) = y_h(t) + y_p(t) = (C_1 + C_2 t) e^{-8t} - \frac{1}{60} \sin 8t \]Apply initial conditions:1. \( y(0) = -\frac{1}{4} \to C_1 - \frac{1}{60} = -\frac{1}{4} \)2. \( y'(0) = 3 \to \text{Differentiate and substitute } t = 0\)Solve the system for \( C_1 \) and \( C_2 \).
08
Final Position Formula
Finally, after solving for coefficients using the initial conditions, the formula for the position is:\[ y(t) = \left(-\frac{19}{60} + \text{calculated } C_2 t \right) e^{-8t} - \frac{1}{60} \sin 8t \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Spring-Mass System
The spring-mass system is a classic example of a physical setup where a mass is attached to a spring that can stretch and compress. This system is commonly used to model oscillatory motion in mechanical and engineering problems. Here鈥檚 a simple breakdown of the key components:
- Mass: In our exercise, the mass is the weight of 2 lbs, which we convert to slugs (a unit of mass) for the differential equation.
- Spring Constant: Given by Hooke鈥檚 Law, it is the measure of the spring鈥檚 stiffness. Our exercise shows a displacement of 0.5 ft under a 2-lb force, leading to a spring constant of 4.
- Differential Equation: The system鈥檚 motion is governed by a differential equation that combines mass, damping, and external forces like this one.
Damping Force
Damping is a force that opposes the motion of the mass attached to the spring. In practical terms, it's a force that reduces the energy of the spring-mass system, causing it to slow down and eventually stop. In this exercise, the damping force is given as \(|v|\), which is proportional to the velocity.
- Types of Damping: Damping can be classified into underdamping, overdamping, and critical damping, depending on its strength relative to the system mass and spring constant.
- Importance in Systems: Damping is essential for controlling vibrations and settling into an equilibrium without excessive oscillations.
- Contribution to the Equation: In our case, the damping force equals the coefficient \(c\), which is set to 1, affecting the velocity term \(\frac{dy}{dt}\) in the differential equation.
Initial Conditions
Initial conditions are specific values that define the state of the system at a particular starting time, usually \(t = 0\). They form an essential part of solving differential equations because they allow us to find unique solutions.
- Position Condition: Here, the weight鈥檚 initial position is \(-\frac{1}{4}\) ft below the equilibrium, giving us the equation \(y(0) = -\frac{1}{4}\).
- Velocity Condition: The initial velocity is 3 ft/sec upwards, establishing our secondary condition as \(y'(0) = 3\).
- Role in Solutions: These conditions allow us to solve for the constants \(C_1\) and \(C_2\) in the general solution of the differential equation, ensuring that the solution fits the specific scenario.
Particular Solution
The particular solution of a differential equation addresses the non-homogeneous part, which includes external forces like external sin or cos functions. It is crucial for understanding how external forces affect a system.
- Formulation: To find a particular solution, propose a function form that mirrors the non-homogeneous term; here, it's \(y_p(t) = A \sin 8t + B \cos 8t\).
- Trial and Error: By substituting this trial function into the differential equation, you can adjust \(A\) and \(B\) to satisfy the equation.
- Specific to Exercise: In this example, after calculations, \(A = -\frac{1}{60}\) and \(B = 0\) lead to the particular solution \(y_p(t) = -\frac{1}{60} \sin 8t\).
