Q39RP - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A 32-lb weight is attached to a vertical spring, causing it to stretch 6 in. upon coming to rest at equilibrium. The damping constant for the system is 2 lb-sec/ft. An external force $\mathbf{f}\left(t\right)\mathbf{=}\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\mathbf{ }\mathbf{lb}$ is applied to the weight. Find the steady-state solution for the system. What is its resonant frequency?

Step-by-Step Solution

Answer

The steady-state solution for the system is $\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}$.

The resonant frequency is;

$\mathbf{f}\mathbf{=}\frac{\sqrt{\mathbf{63}}}{\mathbf{2}\mathbf{\pi }}$

Write the differential equation using the given information

The differential equation is,

$\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\mathbf{=}\mathbf{f}\left(\mathbf{t}\right) ......\left(\mathbf{1}\right)$

From the given information, 

 $$\begin{gathered} \mathbf{m}\mathbf{=}\frac{\mathbf{32}}{\mathbf{32}}\mathbf{=}\mathbf{1} \\ \mathbf{32}\mathbf{=}\mathbf{k}\frac{\mathbf{6}}{\mathbf{12}} \\ \mathbf{k}\mathbf{=}\mathbf{64} \\ \mathbf{b}\mathbf{=}\mathbf{2} \end{gathered}$$

And $\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}$

Substitute the all value of m, k, b and f(t) in the equation (1),

 $\begin{array}{rcl}\mathbf{my}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}& \mathbf{=}& \mathbf{f}\left(\mathbf{t}\right)\\ \left(\mathbf{1}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{2}\right)\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left(\mathbf{64}\right)\mathbf{y}& \mathbf{=}& \mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\\ \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{64}\mathbf{y}& \mathbf{=}& \mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t} ......\left(\mathbf{2}\right)\\ & & \end{array}$

Now find the complimentary solution of the given equation

The auxiliary equation for the above equation,

$\begin{array}{rcl}{\mathbf{m}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{m}\mathbf{+}\mathbf{64}& \mathbf{=}& \mathbf{0}\\ \mathbf{m}& \mathbf{=}& \frac{\mathbf{-}\mathbf{2}\mathbf{\pm }\sqrt{\mathbf{4}\mathbf{-}\mathbf{4}\mathbf{\times }\mathbf{64}}}{\mathbf{2}}\\ \mathbf{m}& \mathbf{=}& \mathbf{-}\mathbf{1}\mathbf{\pm }\mathbf{i}\sqrt{\mathbf{63}}\\ & & \end{array}$

The root of auxiliary equation is, 

${\mathbf{m}}_{\mathbf{1}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{63}}\mathbf{,} {\mathbf{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{63}}$

The complimentary solution of the given equation is,

${\mathbf{y}}_{\mathbf{c}}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right] ......\left(\mathbf{3}\right)$

Find the particular solution to find a general solution for the equation

Assume, the particular solution of equation (1),

${\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\right) ......\left(\mathbf{4}\right)$

Consider,

 $\begin{array}{rcl}\mathbf{X}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\right)\\ \mathbf{Y}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{sin}\mathbf{8}\mathbf{t}\right)\\ & & \\ \mathbf{X}\mathbf{+}\mathbf{iY}& \mathbf{=}& \frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left(\mathbf{4}\mathbf{cos}\mathbf{8}\mathbf{t}\mathbf{+}\mathbf{i}\mathbf{4}\mathbf{sin}\mathbf{8}\mathbf{t}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{D}\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{{\left(\mathbf{8}\mathbf{i}\right)}^{\mathbf{2}}\mathbf{+}\mathbf{2}\left(\mathbf{8}\mathbf{i}\right)\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \mathbf{4}\frac{\mathbf{1}}{\mathbf{-}\mathbf{64}\mathbf{+}\mathbf{16}\mathbf{i}\mathbf{+}\mathbf{64}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \frac{\mathbf{4}}{\mathbf{16}\mathbf{i}}\left({\mathbf{e}}^{\mathbf{8}\mathbf{it}}\right)\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}}\left(\mathbf{sin}\mathbf{8}\mathbf{t}\mathbf{-}\mathbf{icos}\mathbf{8}\mathbf{t}\right)\\ & & \end{array}$

Therefore, the real and imaginary parts are;

$\mathbf{X}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t} and \mathbf{Y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{cos}\mathbf{8}\mathbf{t}$

From the equation (4),

${\mathbf{y}}_{\mathbf{p}}\left(t\right)\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}$

Find the general solution

Therefore, the general solution is,

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{y}}_{\mathbf{c}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{y}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right]\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t} ......\left(\mathbf{5}\right) \end{gathered}$$

Thus, steady state solution of the system for $\mathbf{t}\to \infty$

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\left[\mathbf{Acos}\sqrt{\mathbf{63}}\mathbf{t}\mathbf{+}\mathbf{Bsin}\sqrt{\mathbf{63}}\mathbf{t}\right]\mathbf{+}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t} \\ \mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}\mathbf{,} {\mathbf{e}}^{\mathbf{-}\mathbf{t}}\to \mathbf{0} \mathbf{as} \mathbf{t}\to \infty \end{gathered}$$

Therefore, the solution,

$\mathbf{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{sin}\mathbf{8}\mathbf{t}$

Find the frequency

Thus, the frequency is $\mathbf{f}\mathbf{=}\frac{\sqrt{\mathbf{63}}}{\mathbf{2}\mathbf{\pi }}$.