(a) Show that the general solution of $\frac{d^{2} x}{d t^{2}}+2 \lambda \frac{d x}{d t}+\omega^{2} x=F_{0} \sin \gamma t$ is $x(t)=A e^{-\lambda t} \sin (\sqrt{\omega^{2}-\lambda^{2} t}+\phi)$ $+\frac{F_{0}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4 \lambda^{2} \gamma^{2}}} \sin (\gamma t+\theta)$, where $A=\sqrt{c_{1}^{2}+c_{2}^{2}}$ and the phase angles $\phi$ and $\theta$ are, respectively, defined by $\sin \phi=c_{1} / A, \cos \phi=c_{2} / A$ and $\sin \theta=\frac{-2 \lambda \gamma}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4 \lambda^{2} \gamma^{2}}}$, $\cos \theta=\frac{\omega^{2}-\gamma^{2}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4 \lambda^{2} \gamma^{2}}}$.
(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 $g(y)=\frac{F_{0}}{\sqrt{\left(\omega^{2}-\gamma^{2}\right)^{2}+4 \lambda^{2} \gamma^{2}}}$. Although the amplitude $g(\gamma)$ of $x_{p}(t)$ is bounded as $t \rightarrow \infty$, show that the maximum oscillations will occur at the value $\gamma_{1}=\sqrt{\omega^{2}-2 \lambda^{2}} .$ What is the maximum value of $g ?$ The number $\sqrt{\omega^{2}-2 \lambda^{2} / 2 \pi}$ 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 $\beta \rightarrow 0 ?$