Q13 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In the following problems, take $\mathrm{g}=32\mathrm{ft}/{\sec }^2$ for the U.S. Customary System and $\mathrm{g}=9.8\mathrm{m}/{\sec }^2$ for the MKS system.

A mass weighing 32 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At time t = 0, an external force F(t) = 3 cos 4t lb is applied to the system. If the spring constant is 5 lb/ft and the damping constant is 2 lb-sec/ft, find the steady-state solution for the system.

Step-by-Step Solution

Answer

Therefore, the steady-state solution for the system is:

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{33}{185}\cos 4\mathrm{t}+\frac{24}{185}\sin 4\mathrm{t}$

Step 1: General form

The general solution to (1) in the case $0<{\mathrm{b}}^2<4\mathrm{mk}$:

 

$\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\left(\frac{\mathrm{b}}{2\mathrm{m}}\right)\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}\sin \left(\gamma \mathrm{t}+\mathrm{\theta }\right)$

 

The amplitude of the steady–state solution to equation (1) depends on the angular frequency y of the forcing function and it is given by $\mathrm{A}\left(\gamma \right)={\mathrm{F}}_0\mathrm{M}\left(\gamma \right)$, where

 

(13) $\mathrm{M}\left(\gamma \right):=\frac{1}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}$   … (1)

 

The undamped system:

The system is governed by $\mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}$. And a homogenous solution of it is 

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\mathrm{\omega t}+\mathrm{\varphi }\right),\mathrm{\omega }:=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$. And the corresponding homogeneous equation is 

(21) ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$. And the correct form is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos}\omega \mathrm{t}+{\mathrm{A}}_2\mathrm{tsin}\omega \mathrm{t}$.

So, the general solution of the system is $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

Step 2: Evaluate the equation

Given that, the weight mg = 32 (g = 32). Then, $\mathrm{m}=\frac{32}{32}=1$. And $\mathrm{F}=3\cos 4\mathrm{t}$. So, ${\mathrm{F}}_0=3$and $\gamma \mathbf{=}4$. Since, $\mathrm{k}=5$ and $\mathrm{b}=2$.

 

Then, the differential equation is$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+2\frac{\mathrm{dy}}{\mathrm{dt}}+5\mathrm{y}=3\cos \left(4\mathrm{t}\right)$… (2)

Since one needs to find the steady state solution of the equation. We are only looking for ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)$ Then,

 

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\cos 4\mathrm{t}+{\mathrm{A}}_2\sin 4\mathrm{t}$ … (3)

 

Now find the derivative of equation (3).

 $$\begin{gathered} y_p^{\text{'}}\left(\mathrm{t}\right)=-4{\mathrm{A}}_1\sin 4\mathrm{t}+4{\mathrm{A}}_2\cos 4\mathrm{t} \\ {\mathrm{y}}_{\mathrm{p}}^{\text{'}\text{'}}\left(\mathrm{t}\right)=-16{\mathrm{A}}_1\cos 4\mathrm{t}-16{\mathrm{A}}_2\sin 4\mathrm{t} \end{gathered}$$

Substitute the values in equation (2).

 

$$\begin{gathered} -16{\mathrm{A}}_1\cos 4\mathrm{t}-16{\mathrm{A}}_2\sin 4\mathrm{t}-8{\mathrm{A}}_1\sin 4\mathrm{t}+8{\mathrm{A}}_2\cos 4\mathrm{t}+5{\mathrm{A}}_1\cos 4\mathrm{t}+{\mathrm{A}}_2\sin 4\mathrm{t}=3\cos \left(4\mathrm{t}\right) \\ \left(-11{\mathrm{A}}_1+8{\mathrm{A}}_2\right)\cos \left(4\mathrm{t}\right)+\left(-8{\mathrm{A}}_1-11{\mathrm{A}}_2\right)\sin \left(4\mathrm{t}\right)=3\cos \left(4\mathrm{t}\right) \end{gathered}$$

 

Now equalize the like terms.

 $$\begin{gathered} -11{\mathrm{A}}_1+8{\mathrm{A}}_2=3 \\ -8{\mathrm{A}}_1-11{\mathrm{A}}_2=0 \\ {\mathrm{A}}_2=-\frac{8{\mathrm{A}}_1}{11} \end{gathered}$$

Then,

 $$\begin{gathered} -11{\mathrm{A}}_1-8\times \frac{8{\mathrm{A}}_1}{11}=3 \\ -121{\mathrm{A}}_1-64{\mathrm{A}}_1=33 \\ -185{\mathrm{A}}_1=33 \\ {\mathrm{A}}_1=-\frac{33}{185} \end{gathered}$$

 And

$$\begin{gathered} {\mathrm{A}}_2=-\frac{8\times \frac{-33}{185}}{11} \\ =\frac{8\times 33}{11\times 185} \\ =\frac{24}{185} \end{gathered}$$

The solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{33}{185}\cos 4\mathrm{t}+\frac{24}{185}\sin 4\mathrm{t}$.