(a) Show that the general solution of

Chapter 5, Problem 43E

(choose chapter or problem)

(a) Show that the general solution of

\(\frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+\omega^{2} x=F_{0} \sin \gamma t\)

is

\(x(t)=A e^{-\lambda t} \sin \left(\sqrt{\omega^{2}-\lambda^{2}} t+\phi\right)\)

\(+\frac{F_{0}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4^{2} \gamma^{2}}} \sin (\gamma t+\theta)\)

where \(A=\sqrt{c_{1}^{2}+c_{2}^{2}}\) and the phase angles \(\phi\) and \(\boldsymbol{\theta}\) are, respectively, defined by sin \(\phi=c_{1} / A\), \(\cos \phi=c_{2} / A\) and

\(\begin{array}{l}\sin \theta=\frac{-2 \gamma}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4{ }^{2} \gamma^{2}}} \\

\cos \theta=\frac{\omega^{2}-\gamma^{2}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4^{2} \gamma^{2}}}\end{array}\)

(b) The solution in part (a) has the form \(x(t)=x_{c}(t)+x_{p}(t)\). Inspection shows that \(x_{c}(t)\) is transient, and hence for large values of time, the solution is approximated by \(x_{p}(t)=g(\gamma) \sin (\gamma t+\theta)\), where

\(g(y)=\frac{F_{0}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4{ }^{2} \gamma^{2}}}\).

Although the amplitude \(g(\gamma)\) of \(x_{p}(t)\) is bounded as \(t \longrightarrow \infty\), show that the maximum oscillations will occur at the value \(\gamma_{1}=\sqrt{\omega^{2}-2^{2}}\). What is the maximum value of g? The number \(\sqrt{\omega^{2}-22} / 2 \pi\) is said to be the resonance frequency of the system.

(c) When \(F_{0}=2, m=1\), and k = 4 g becomes

\(g(\gamma)=\frac{2}{\sqrt{\left(4-\gamma^{2}\right)^{2}+\beta^{2} \gamma^{2}}}\)

Construct a table of the values of \(\gamma_{1}\) and \(g\left(\gamma_{1}\right)\) corresponding to the damping coefficients \(\beta=2, \beta=1\), \(\beta=\frac{3}{4}, \beta=\frac{1}{2}\), and \(\beta=\frac{1}{4}\). Use a graphing utility to obtain the graphs of g corresponding to these damping coefficients. Use the same coordinate axes. This family of graphs is called the resonance curve or frequency response curve of the system. What is \(\gamma_{1}\), approaching as \(\beta \rightarrow 0\)? What is happening to the resonance curve as \(\beta \rightarrow 0\)?

Text Transcription:

fracd^2xdt^2+2fracdxdt+omega^2x=F_0\sin\gammat

x(t)=Ae^-lambdat\sin(sqrtomega^2-lambda^2t+phi)

+fracF_0sqrtt(omega^2-gamma^2)^2+4^2\gamma^2sin(gammat+theta)

A=sqrtc_1^2+c_2^2

phi

theta

phi=c_1/A

cos\phi=c_2/A

l\sin\theta=frac-2\gammasqrt(omega^2-gamma^2)^2+4^2gamma^2

F_0=2,m=1

g(gamma)=frac2sqrt(4-gamma^2)^2+beta^2\gamma^2

cos\theta=fracomega^2-gamma^2\sqrt(omega^2-gamma^2^2+4^2gamma^2xt=x_c(t)+x_p(t)

x_c(t)

x_p}(t)=g(gamma)sin(gammat+theta)

g(y)=fracF_0\sqrt(omega^2-gamma^2)^2+4^2gamma^2

g(gamma)

x_p(t)

t\infty

gamma_1=sqrtomega^2-2^2

sqrt\omega^2-22/2pi

gamma_1

g(gamma_1

beta=2,beta=1

beta=frac34,beta=frac12

beta=frac14

beta0