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Chapter 10: Problem 67

Determine the values of the appropriate parameters needed to give the systems governed by the following second-order linear constant-coefficient differential equations the damping parameters and natural frequencies stated: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 a \frac{\mathrm{d} x}{\mathrm{~d} t}+b x=0, \quad \zeta=0.5, \quad \omega=2\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+p \frac{\mathrm{d} x}{\mathrm{~d} t}+q x=0, \quad \zeta=1.4, \quad \omega=0.5\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\beta \frac{\mathrm{d} x}{\mathrm{~d} t}+\gamma x=0, \quad \zeta=1, \quad \omega=1.1\)

Short Answer

Expert verified
(a) \(a=1, b=4\); (b) \(p=1.4, q=0.25\); (c) \(\beta=2.2, \gamma=1.21\).

Step by step solution

01

Understand the Relationship

For a second-order differential equation \( \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}} + 2\zeta\omega_n \frac{\mathrm{d} x}{\mathrm{d} t} + \omega_n^2 x = 0 \), the damping ratio \( \zeta \) and natural frequency \( \omega_n \) can be related to the equation's parameters. Coefficients before \( \frac{\mathrm{d} x}{\mathrm{d} t} \) and \( x \) are \( 2\zeta\omega_n \) and \( \omega_n^2 \), respectively.
02

Solve for (a) Parameters

Given \( \zeta = 0.5 \) and \( \omega = 2 \), substitute into \( 2a = 2\zeta\omega_n \) and \( b = \omega_n^2 \). Thus, \( 2a = 2\times0.5\times2 = 2 \) giving \( a = 1 \), and \( b = (2)^2 = 4 \).
03

Solve for (b) Parameters

Given \( \zeta = 1.4 \) and \( \omega = 0.5 \), substitute into \( p = 2\zeta\omega_n \) and \( q = \omega_n^2 \). Thus, \( p = 2\times1.4\times0.5 = 1.4 \), and \( q = (0.5)^2 = 0.25 \).
04

Solve for (c) Parameters

Given \( \zeta = 1 \) and \( \omega = 1.1 \), substitute into \( \beta = 2\zeta\omega_n \) and \( \gamma = \omega_n^2 \). Thus, \( \beta = 2\times1\times1.1 = 2.2 \), and \( \gamma = (1.1)^2 = 1.21 \).

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Key Concepts

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

Damping Ratio
The damping ratio \( \zeta \) is a dimensionless measure describing how oscillations in a system decay after a disturbance. In the context of second-order differential equations, it's crucial in determining the behavior of systems described by equations like \( \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}} + 2\zeta\omega_n \frac{\mathrm{d} x}{\mathrm{d} t} + \omega_n^2 x = 0 \). Here, \( \zeta \) affects whether the system experiences overdamped, critically damped, or underdamped oscillations.

  • If \( \zeta > 1 \), the system is overdamped, meaning it returns to equilibrium without oscillating.
  • When \( \zeta = 1 \), the system is critically damped, allowing it to return to equilibrium as quickly as possible without oscillating.
  • For \( \zeta < 1 \), the system is underdamped, causing it to oscillate before settling to equilibrium.
The damping ratio is crucial in engineering to design systems that do not oscillate excessively or take too long to stabilize.
Natural Frequency
The natural frequency \( \omega_n \) is the frequency at which a system oscillates when not subjected to any external force or damping. It is an inherent property of the system, related to its mass and stiffness. Mathematically, it appears in the standard form of the second-order differential equation as \( \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}} + 2\zeta\omega_n \frac{\mathrm{d} x}{\mathrm{d} t} + \omega_n^2 x = 0 \).

  • In mechanical systems, \( \omega_n \) depends on physical parameters like mass and damping characteristics.
  • Electrical circuits also exhibit a natural frequency determined by inductance and capacitance values.
Understanding the natural frequency helps predict how a system might behave under various conditions or when designed for specific functional requirements.
Linear Constant-Coefficient Differential Equations
Linear constant-coefficient differential equations are integral in modeling a wide range of physical systems. These equations have the form \( a_n \frac{\mathrm{d}^{n} x}{\mathrm{d} t^{n}} + a_{n-1} \frac{\mathrm{d}^{n-1} x}{\mathrm{d} t^{n-1}} + \, ... \, + a_0 x = 0 \), where the coefficients \( a_n, a_{n-1}, ..., a_0 \) are constants.

Especially in second-order equations like \( \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}} + 2\zeta\omega_n \frac{\mathrm{d} x}{\mathrm{d} t} + \omega_n^2 x = 0 \), these constant coefficients make them comparatively simple to solve analytically.

  • Such equations are used to describe the dynamic behavior of systems across disciplines, including mechanical vibrations, electrical circuits, and control systems.
  • The assumption of constant coefficients simplifies the interpretation and solution of the equations, assuming system parameters remain constant over time.
This type of equation is fundamental in engineering and physics, providing insights into system stability and response to inputs.

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Most popular questions from this chapter

Find the general solutions of the following differential equations: (a) \(x t \frac{\mathrm{d} x}{\mathrm{~d} t}=x^{2}+t^{2}\) (b) \(x^{2} \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{3}+x^{3}}{t}\) (c) \(t \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x^{2}+x t}{t}\)

For each of the following sets of linearly dependent functions find \(k_{1}, k_{2}, \ldots\) such that \(k_{1} f_{1}+k_{2} f_{2}+\ldots=0\) (a) \(\\{\sin t, \cos t+\sin t, \cos 2 t-\sin t, \cos t-\cos 2 t\\}\) (b) \(\left\\{t+t^{3}, t-t^{2}, t^{2}+2 t^{3}, t^{2}-t^{3}\right\\}\) (c) \(\left\\{\ln t, \ln 2 t, \ln 4 t^{2}\right\\}\) (d) \(\\{f(t)+g(t), f(t)(1+f(t)), g(t)-f(t)\) \(\left.f(t)^{2}-g(t)\right\\}\) (e) \(\left\\{1+t+2 t^{2}, t-2 t^{2}+3 t^{3}, 1+t-2 t^{2},\right.\), \(\left.t-2 t^{2}-3 t^{3}, t^{3}\right\\}\)

State which of the following problems are under-determined (that is, have insufficient boundary conditions to determine all the arbitrary constants in the general solution) and which are fully determined. In the case of fully determined problems state which are boundary-value problems and which are initial-value problems. (Do not attempt to solve the differential equations.) (a) \(4 x \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\left(2 t^{2}-\frac{1}{x}\right) \frac{\mathrm{d} x}{\mathrm{~d} t}-4 x^{2} t=0, \quad x(0)=4\) (b) \(\left(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}\right)^{2}+t \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-x\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}=0\) $$ x(0)=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(0)=1, \quad x(2)=0 $$ (c) \(\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}-x^{2}=\sin t, \quad x(0)=a\) (d) \(\frac{\mathrm{d}^{4} x}{\mathrm{~d} t^{4}}+4 \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-2 \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}-4 x=\mathrm{e}^{t}\) \(x(0)=1, \quad x(2)=0\) (e) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-2 t \frac{\mathrm{d} x}{\mathrm{~d} t}=t^{2}-4, \quad x(0)=1, \quad x(2)=0\) (f) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 x\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}-\frac{x}{t}=0\) $$ x(1)=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(1)=4 $$ (g) \(\left(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}\right)^{2}+2 t=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(2)=1\) (h) \(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}} \frac{\mathrm{d} x}{\mathrm{~d} t}+x \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=2 t^{2}\) $$ x(0)=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(0)=0 $$ (i) \(\left(\frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}\right)^{1 / 2}+t \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+x \frac{\mathrm{d} x}{\mathrm{~d} t}-\frac{x}{t}=0\) $$ x(1)=1, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(1)=0, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(3)=0 $$ (j) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=(x-t)^{2}, \quad x(4)=2\) (k) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}-4 \frac{\mathrm{d} x}{\mathrm{~d} t}+4 x=\cos t, \quad x(1)=0, \quad x(3)=0\) (1) \(\frac{1}{t} \frac{\mathrm{d}^{3} x}{\mathrm{~d} t^{3}}-t^{2}\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{2}+x\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)^{1 / 2}-\left(t^{2}+4\right) x=0\) $$ x(0)=0, \quad \frac{\mathrm{d} x}{\mathrm{~d} t}(0)=U, \quad \frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}(0)=0 $$

Sketch the direction field of the differential equation $$ \frac{\mathrm{d} x}{\mathrm{~d} t}=-2 t $$ Find the solution of the equation. Sketch the particular solutions for which \(x(0)=2\), and for which \(x(2)=-3\), and check that these are consistent with your direction field.

Draw phase-plane direction fields for the following equations and sketch the form you would expect the solution paths to take, starting from the points \((x, v)=(1,0),(0,1),(-1,0)\) and \((0,-1)\) in each case: (a) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}+x^{3}=0\) (b) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}+\operatorname{sgn}(x)=0\) (c) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\frac{\mathrm{d} x}{\mathrm{~d} t}+x^{2} \operatorname{sgn}(x)=0\) (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+\operatorname{sgn}\left(\frac{\mathrm{d} x}{\mathrm{~d} t}\right)+2 \operatorname{sgn}(x)=0\)
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