91Ó°ÊÓ

The initial-value problem is \\[ x^{\prime \prime}+\frac{2}{m} x^{\prime}+\frac{k}{m} x=0, \quad x(0)=0, x^{\prime}(0)=v_{0} \\] With \(k=10,\) the auxiliary equation has roots \(\gamma=-1 / m \pm \sqrt{1-10 m} / m .\) Consider the three cases: \((i) m=\frac{1}{10} .\) The roots are \(\gamma_{1}=\gamma_{2}=10\) and the solution of the differential equation is \(x(t)=c_{1} e^{-10 t}+c_{2} t e^{-10 t}\) The initial conditions imply \(c_{1}=0\) and \(c_{2}=v_{0}\) and so \(x(t)=v_{0} t e^{-10 t} .\) The condition \(x(1)=0\) implies \(v_{0} e^{-10}=0\) which is impossible because \(v_{0} \neq 0\) \((i i) 1-10 m>0\) or 0 < m < \frac{1}{10}.\( \)The roots are $$\gamma_{1}= -\frac{1}{m}-\frac{1}{m} \sqrt{1-10 m} \quad \text { and }$$ $$\gamma_{2}=-\frac{1}{m}+\frac{1}{m} \sqrt{1-10 m}$$ and the solution of the differential equation is \(x(t)=c_{1} e^{\gamma_{1} t}+c_{2} e^{\gamma_{2} t}\). The initial conditions imply $$\begin{array}{r} c_{1}+c_{2}=0 \\ \gamma_{1} c_{1}+\gamma_{2} c_{2}=v_{0} \end{array}$$ so \(c_{1}=v_{0} /\left(\gamma_{1}-\gamma_{2}\right), c_{2}=-v_{0} /\left(\gamma_{1}-\gamma_{2}\right),\) and $$x(t)=\frac{v_{0}}{\gamma_{1}-\gamma_{2}}\left(e^{\gamma_{1} t}-e^{\gamma_{2} t}\right)$$ Again, \(x(1)=0\) is impossible because \(v_{0} \neq 0\). \((i i i) \quad 1-10 m < 0\) or \(m >\frac{1}{10} .\) The roots of the auxiliary equation are $$\gamma_{1}=-\frac{1}{m}-\frac{1}{m} \sqrt{10 m-1} i \quad \text { and }\space\gamma_{2}=-\frac{1}{m}+\frac{1}{m} \sqrt{10 m-1}$$ and the solution of the differential equation is $$x(t)=c_{1} e^{-t / m} \cos \frac{1}{m} \sqrt{10 m-1} t+c_{2} e^{-t / m} \sin \frac{1}{m} \sqrt{10 m-1} t$$ The initial conditions imply \(c_{1}=0\) and \(c_{2}=m v_{0} / \sqrt{10 m-1},\) so that $$x(t)=\frac{m v_{0}}{\sqrt{10 m-1}} e^{-t / m} \sin \left(\frac{1}{m} \sqrt{10 m-1} t\right)$$ The condition \(x(1)=0\) implies $$\begin{aligned} \frac{m v_{0}}{\sqrt{10 m-1}} e^{-1 / m} \sin \frac{1}{m} \sqrt{10 m-1} &=0 \\\ \sin \frac{1}{m} \sqrt{10 m-1} &=0 \\ \frac{1}{m} \sqrt{10 m-1} &=n \pi \\ \frac{10 m-1}{m^{2}} &=n^{2} \pi^{2}, n=1,2,3, \ldots \\ \left(n^{2} \pi^{2}\right) m^{2}-10 m+1 &=0 \end{aligned}$$ $$m=\frac{10 \sqrt{100-4 n^{2} \pi^{2}}}{2 n^{2} \pi^{2}}=\frac{5 \pm \sqrt{25-n^{2} \pi^{2}}}{n^{2} \pi^{2}}$$ since \(m\) is real, \(25-n^{2} \pi^{2} \geq 0 .\) If \(25-n^{2} \pi^{2}=0,\) then \(n^{2}=25 / \pi^{2},\) and \(n\) is not an integer. Thus, \(25-n^{2} \pi^{2}=\) \((5-n \pi)(5+n \pi) >0\) and since \(n >0,5+n \pi >0,\) so \(5-n \pi >0\) also. Then \(n<5 / \pi,\) and so \(n=1 .\) Therefore the mass \(m\) will pass through the equilibrium position when \(t=1\) for $$m_{1}=\frac{5+\sqrt{25-\pi^{2}}}{\pi^{2}} \quad \text { and } \quad m_{2}=\frac{5-\sqrt{25-\pi^{2}}}{\pi^{2}}$$.

Step by step solution

01

Analyze Initial Value Problem

The initial-value problem is given by the second-order linear differential equation \(x'' + \frac{2}{m} x' + \frac{k}{m} x = 0\) with initial conditions \(x(0) = 0\) and \(x'(0) = v_0\). For the given case, \(k = 10\), and the auxiliary equation is solved to find roots that determine the nature of the solutions.

02

Solve for Case (i) \(m = \frac{1}{10}\)

For the case \(m = \frac{1}{10}\), the auxiliary equation yields roots \(\gamma_1 = \gamma_2 = 10\). The solution to the differential equation is \(x(t) = c_1 e^{-10t} + c_2 t e^{-10t}\). Given the initial conditions \(x(0) = 0\) and \(x'(0) = v_0\), we determine \(c_1 = 0\) and \(c_2 = v_0\), leading to \(x(t) = v_0 t e^{-10t}\). The condition \(x(1) = 0\) implies \(v_0 e^{-10} = 0\), which is impossible since \(v_0 eq 0\). Thus, no valid solution for \(x(t)\) under these conditions exists.

03

Analyze Case (ii) \(0 < m < \frac{1}{10}\)

In this case, the inequality \(1 - 10m > 0\) is satisfied, and the auxiliary equation yields distinct real roots \(\gamma_1 = -\frac{1}{m} - \frac{\sqrt{1-10m}}{m}\) and \(\gamma_2 = -\frac{1}{m} + \frac{\sqrt{1-10m}}{m}\). The general solution is \(x(t) = c_1 e^{\gamma_1 t} + c_2 e^{\gamma_2 t}\). From the initial conditions, we find \(c_1 = \frac{v_0}{\gamma_1 - \gamma_2}\) and \(c_2 = -\frac{v_0}{\gamma_1 - \gamma_2}\), yielding \(x(t) = \frac{v_0}{\gamma_1 - \gamma_2}(e^{\gamma_1 t} - e^{\gamma_2 t})\). The condition \(x(1) = 0\) is again impossible to satisfy since \(v_0 eq 0\).

04

Consider Case (iii) \(m > \frac{1}{10}\)

For \(m > \frac{1}{10}\), the term \(1 - 10m < 0\), which results in complex roots \(\gamma_1 = -\frac{1}{m} - i \frac{\sqrt{10m-1}}{m}\) and \(\gamma_2 = -\frac{1}{m} + i \frac{\sqrt{10m-1}}{m}\). This leads to the solution \(x(t) = c_1 e^{-t/m} \cos\left(\frac{\sqrt{10m-1}}{m} t\right) + c_2 e^{-t/m} \sin\left(\frac{\sqrt{10m-1}}{m} t\right)\). Using initial conditions \(x(0) = 0\) and \(x'(0) = v_0\), find \(c_1 = 0\) and \(c_2 = \frac{m v_0}{\sqrt{10m-1}}\). Thus, \(x(t) = \frac{m v_0}{\sqrt{10m-1}} e^{-t/m} \sin\left(\frac{\sqrt{10m-1}}{m} t\right)\).

05

Derive Mass Values for Equilibrium at \(t = 1\)

The condition \(x(1) = 0\) leads to \(e^{-1/m} \sin\left(\frac{\sqrt{10m-1}}{m}\right) = 0\), implying \(\sin\left(\frac{\sqrt{10m-1}}{m}\right) = 0\), or \(\frac{\sqrt{10m-1}}{m} = n\pi\), where \(n = 1, 2, 3, \ldots\). Solving for \(m\) gives the quadratic equation \((n^2\pi^2)m^2 - 10m + 1 = 0\), which has solutions \(m = \frac{5 \pm \sqrt{25 - n^2 \pi^2}}{n^2 \pi^2}\). For \(m\) to be real, ensure \(25 - n^2 \pi^2 \geq 0\). This constraint limits \(n\) so that \(n < 5/\pi\), with only \(n = 1\) satisfying this condition. Hence, possible mass values are \(m_1 = \frac{5 + \sqrt{25 - \pi^2}}{\pi^2}\) and \(m_2 = \frac{5 - \sqrt{25 - \pi^2}}{\pi^2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differential Equations

Differential equations are equations that include a function and its derivatives. They are crucial tools in modeling situations where change is continuous. In the context of an initial-value problem, we solve a differential equation with given conditions at a specific point, often when time is zero. This helps us predict future behavior of a system. The differential equation presented in the exercise, \( x'' + \frac{2}{m} x' + \frac{k}{m} x = 0 \), is a second-order linear differential equation. Here, \( x'' \) represents the acceleration or second derivative of the function \( x \), and \( x' \) is the velocity, or first derivative. Understanding these equations allows us to solve problems related to mechanics, such as the motion of objects.

Dynamics and Motion

Dynamics is the study of the forces and motion they produce. In this exercise, dynamics is represented through a differential equation that models an object's movement over time. This particular problem depicts a physical system where a mass moves under certain forces, characterized by the mass \( m \), the damping factor \( \frac{2}{m} \), and a stiffness constant \( \frac{k}{m} \).

By analyzing how the system behaves given initial conditions like starting position and velocity, we can predict its motion. Specifically, this involves looking at different scenarios based on the mass \( m \) and understanding which conditions lead to stable motion. This is fundamental in engineering and physics when studying object dynamics under forces.

Solutions of Second-order Linear Differential Equations

Second-order linear differential equations are key in analyzing systems with dynamic behavior, especially those related to motion and oscillations. These equations typically involve constants and are characterized by their degree, which in this case is two.

Solving them involves finding the roots of the auxiliary (or characteristic) equation, which gives insight into the behavior of the system. Depending on the discriminant of the quadratic in the auxiliary equation, the roots could be real and distinct, real and equal, or complex conjugates. Each case presents a different type of solution involving exponential, oscillatory, or overdamped responses.

In scenarios where the initial conditions are provided, such as in the exercise, these solutions are further defined by constants determined through those conditions. This results in a complete solution that not only satisfies the differential equation but also aligns with the physical starting point of the system.