/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Suppose that an object of mass \... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 22

Suppose that an object of mass \(m=1\) slug is attached to a spring with spring constant \(k=25 \mathrm{lb} / \mathrm{ft}\). If the resistive force is \(F_{R}=\) \(6 d x / d t\), determine the displacement of the object if it is set into motion from its equilibrium position with an upward velocity of \(2 \mathrm{ft} / \mathrm{s}\). What is the quasiperiod of the motion?

Short Answer

Expert verified
The displacement of the object is: \( x(t) = \frac{1}{2} e^{-3t} \sin(4t) \). The quasiperiod is \( \frac{\pi}{2} \).

Step by step solution

01

- Write down the differential equation

The differential equation for damped motion is given by \[ m\frac{d^2 x}{dt^2} + c\frac{dx}{dt} + kx = 0 \]Given: m = 1 \text{ slug}, c = 6 \text{ lb路s/ft}, k = 25 \text{ lb/ft}.Thus, the differential equation becomes:\[ \frac{d^2 x}{dt^2} + 6\frac{dx}{dt} + 25x = 0 \]
02

- Solve the characteristic equation

The characteristic equation associated with the differential equation is:\[ r^2 + 6r + 25 = 0 \]Solve for the roots using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \[ a = 1, \ b = 6 , \ c = 25 \].Calculate the discriminant: \[ 6^2 - 4 \cdot 1 \cdot 25 = 36 - 100 = -64 \].Thus, the roots are:\[ r = \frac{-6 \pm \sqrt{-64}}{2} = \frac{-6 \pm 8i}{2} = -3 \pm 4i \].
03

- Form the general solution

For complex roots \( r = \alpha \pm i \beta \), the general solution is:\[ x(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \],where \( \alpha = -3 \) and \( \beta = 4 \).Substitute the values:\[ x(t) = e^{-3t}(C_1 \cos(4t) + C_2 \sin(4t)) \].
04

- Apply initial conditions

Initial conditions are given as \( x(0) = 0 \) and \( \frac{dx}{dt}|_{t=0} = 2 \text{ ft/s} \). Using \( x(0) = 0 \):\[ 0 = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 \Rightarrow C_1 = 0 \].Using \( \frac{dx}{dt}|_{t=0} = 2 \text{ ft/s} \): \[ x'(t) = \frac{d}{dt} \left( e^{-3t}(C_1 \cos(4t) + C_2 \sin(4t)) \right) \].Upon differentiation and applying initial conditions:\[ x'(t) = e^{-3t} \left(C_2 \cdot 4 \cos(4t) \right) - 3e^{-3t}(C_2 \sin(4t)) \].At \( t = 0 \): \[ 2 = 4C_2 \quad\Rightarrow\quad C_2 = \frac{1}{2} \].Thus the displacement is:\[ x(t) = e^{-3t} \left( \frac{1}{2} \sin(4t) \right) \].
05

- Determine the quasiperiod

The quasiperiod is calculated using the imaginary part of the roots, \( \beta = 4 \). The quasiperiod \( T \) is:\[ T = \frac{2\pi}{\beta} = \frac{2\pi}{4} = \frac{\pi}{2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve functions and their derivatives. In the context of damped oscillations, they are used to model the motion of systems where a mass, attached to a spring, experiences a resistive force.
For our problem, the differential equation describing the damped motion is: \[ m\frac{d^2 x}{dt^2} + c\frac{dx}{dt} + kx = 0 \] Here:
  • \(m\) is the mass.
  • \(c\) is the damping coefficient.
  • \(k\) is the spring constant.
Inserting the given values results in: \[ \frac{d^2 x}{dt^2} + 6\frac{dx}{dt} + 25x = 0 \] This form simplifies the process for solving how the mass moves over time.
Characteristic Equation
The characteristic equation is a key step in solving differential equations. It helps us find the nature of the solution.
From our differential equation, \( \frac{d^2 x}{dt^2} + 6\frac{dx}{dt} + 25x = 0 \), we derive the characteristic equation by assuming solutions of the form \( x(t) = e^{rt} \).
This gives us the quadratic equation: \[ r^2 + 6r + 25 = 0 \] Applying the quadratic formula, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 6 \), and \( c = 25 \).
The discriminant \( 36 - 100 = -64 \) reveals complex roots: \[ r = \frac{-6 \pm 8i}{2} = -3 \pm 4i \] These roots guide the form of our solution.
Quasiperiod
The concept of the quasiperiod is important in understanding the behavior of damped oscillations, especially when dealing with complex roots.
In the solution we derive: \( -3 \pm 4i \), the real part \( \alpha = -3 \) and the imaginary part \( \beta = 4 \). The quasiperiod \( T \) is determined using the formula:
\[ T = \frac{2\pi}{\beta} \]
For our values:
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \]
This means the oscillation has a repeating pattern every \( \frac{\pi}{2} \) seconds, but with diminishing amplitude due to the damping effect.
Initial Conditions
Initial conditions are essential for determining the specific solution to our differential equation. They tell us the state of the system at time \( t = 0 \).
In this problem, the initial conditions are given as:
  • \( x(0) = 0 \)
  • \( \frac{dx}{dt}|_{t=0} = 2 \text{ ft/s} \)
Using \( x(0) = 0 \), we set up the equation:
\(C_1 \cos(0) + C_2 \sin(0) = 0 \rightarrow C_1 = 0\)
Using \( \frac{dx}{dt}|_{t=0} = 2 \text{ ft/s} \), we differentiate and solve:
\[ x'(t) = e^{-3t} \left(C_2 \cdot 4 \cos(4t) \right) - 3e^{-3t}(C_2 \sin(4t)) \]
At \( t = 0 \),
\[ 2 = 4C_2 \rightarrow C_2 = \frac{1}{2} \]
This gives us the specific solution for the displacement of the mass over time.
Mass-Spring Systems
A mass-spring system consists of a mass attached to a spring, which can oscillate back and forth. The dynamics of such a system are described using Hooke's Law and Newton's second law.
For our system:
  • Mass \(m = 1 \text{ slug} \)
  • Spring constant \(k = 25 \text{ lb/ft} \)
  • Damping coefficient \(c = 6 \text{ lb路s/ft} \)
The differential equation governing its motion, \( \frac{d^2 x}{dt^2} + 6\frac{dx}{dt} + 25x = 0 \), models how the mass moves over time under the influence of the spring force and damping.
The solution to this equation gives us insight into the behavior of the mass, showing it undergoes damped oscillations with a specific quasiperiod and exponentially decreasing amplitude.

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Most popular questions from this chapter

Consider Van-der-Pol's equation, \(\frac{d^{2} x}{d t^{2}}+\) \(\epsilon\left(x^{2}-1\right) \frac{d x}{d t}+x=0\), where \(\epsilon\) is a small positive number. Notice that this equation has a nonconstant damping coefficient. Because \(\epsilon\) is small, we can approximate a solution of Van-der-Pol's equation with \(x(t)=A \cos \omega t\), a solution of \(d^{2} x / d t^{2}+x=0\) (the equation obtained when \(\epsilon=0\) ) if \(\omega=1\). This method of approximation is called harmonic balance. (a) Substitute \(x(t)=A \cos \omega t\) into the nonlinear term in Van-der-Pol's equation to obtain \(\epsilon\left(x^{2}-1\right) \frac{d x}{d t}=-\epsilon A \omega\left(\frac{1}{4} A^{2}-1\right) \sin \omega t\) $$ -\frac{1}{4} \epsilon A^{4} \omega \sin 3 \omega t . $$ (b) If we ignore the term involving the higher harmonic \(\sin 3 \omega t\), we have \(\epsilon\left(x^{2}-1\right) d x / d t=-\epsilon A \omega\left(\frac{1}{4} A^{2}-1\right) \sin \omega t\) \(=\epsilon\left(\frac{1}{4} A^{2}-1\right) d x / d t\). Substitute this expression into Van-der-Pol's equation to obtain the linear equation \(d^{2} x / d t^{2}+\epsilon\left(\frac{1}{4} A^{2}-1\right) d x / d t+x=0 .\) (c) If \(A=2\) in the linear equation in (b), is the approximate solution periodic? (d) If \(A \neq 2\) in the linear equation in (b), is the approximate solution periodic?

Solve the initial value problem \(x^{\prime \prime}+x=\) \(\cos \omega t, x(0)=0, x^{\prime}(0)=1\) using values \(\omega=0.9\) and \(\omega=0.7\). Graph the solutions simultaneously to determine the effect that the value of \(\omega\) has on each solution.

39\. Solve the initial value problem \(x^{\prime \prime}+c x^{\prime}+\) \(x=0, x(0)=-1, x^{\prime}(0)=3\) for values of \(c=2\) and \(c=\sqrt{8}\). (Notice the effect of the coefficient of damping on the solution to the corresponding critically damped and overdamped equation.)

To hear resonance, solve the initial value problem \(d^{2} x / d t^{2}+\omega^{2} x=F \cos \omega t, x(0)=0\), \(d x / d t(0)=0\) using \(\omega^{2}=6000\) and \(F=2\). In each case, plot and, if possible, play the solution. (Note: The purpose of the high frequencies is to assist in hearing the solutions when they are played.)

Solve the initial value problem \(x^{\prime \prime}+0.1 x^{\prime}+\) \(x=3 \cos 2 t, x(0)=0, x^{\prime}(0)=0\), using the method of undetermined coefficients and compare the result with the forcing function \(3 \cos 2 t\). Determine the phase difference between these two functions.
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