Q11 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A mass weighing 8 lb is attached to a spring hanging from the ceiling and comes to rest at its equilibrium position. At t = 0, an external force F(t) = 2 cos 2t lb is applied to the system. If the spring constant is 10 lb/ft and the damping constant is 1 lb-sec/ft, find the equation of motion of the mass. What is the resonance frequency for the system?

Step-by-Step Solution

Answer

Therefore, the equation of motion of the mass is:

   $\mathrm{y}\left(\mathrm{t}\right)=-\frac{2\sqrt{34}}{51}{\mathrm{e}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+{\tan }^{-1}\left(\frac{27}{11}\right)\right)+\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+{\tan }^{-1}\left(\frac{9}{2}\right)\right)$

And the resonance frequency for the system is $\frac{2\sqrt{2}}{\mathrm{\pi }}$ .

 

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 angular frequency:

 

The amplitude of the steady-state solution to equation (1) depends on the angular frequency   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 the homogenous solution of it is 

${\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right),\omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ . 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  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_1\mathrm{tcos\omega t}+{\mathrm{A}}_2\mathrm{tsin\omega t}$.

 

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 = 8 (g = 32). Then,

$$\begin{gathered} \mathrm{m}=\frac{8}{32} \\ =\frac{1}{4} \end{gathered}$$

And $\mathrm{F}=2\cos 2\mathrm{t}$ . So, ${\mathrm{F}}_0=2$  and $\gamma =2$ . Since,  $\mathrm{k}=10$ and $\mathrm{b}=1$ .

Then, the differential equation is  $\frac{1}{4}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\frac{\mathrm{dy}}{\mathrm{dt}}+10\mathrm{y}=2\cos \left(2\mathrm{t}\right)$

Since, 

$$\begin{gathered} {\mathrm{b}}^2=1 \\ 4\mathrm{mk}=10 \end{gathered}$$

One knows that, ${\mathrm{b}}^2<4\mathrm{mk}$   so we are dealing with the underdamped motion.

 

The homogeneous solution is given by

$$\begin{gathered} {\mathrm{y}}_{\mathrm{h}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\frac{\mathrm{b}}{2\mathrm{m}}\mathrm{t}}\sin \left(\frac{\sqrt{4\mathrm{mk}-{\mathrm{b}}^2}}{2\mathrm{m}}\mathrm{t}+\varphi \right) \\ ={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(\frac{\sqrt{10-1}}{\frac{1}{2}}\mathrm{t}+\varphi \right) \\ ={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\varphi \right) \end{gathered}$$

Where A and $\varphi$  are constants.

 

Then, the particular solution is given by

 

 $$\begin{gathered} {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{\sqrt{{\left(\mathrm{k}-\mathrm{m}{\gamma }^2\right)}^2+{\mathrm{b}}^2{\gamma }^2}}\sin \left(\gamma \mathrm{t}+\mathrm{\theta }\right) \\ =\frac{2}{\sqrt{{\left(10-1\right)}^2+4}}\sin \left(2\mathrm{t}+\mathrm{\theta }\right) \\ =\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+\mathrm{\theta }\right) \end{gathered}$$

 The general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\varphi \right)+\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+\mathrm{\theta }\right)$   … (2)

Step 3: Implement the initial condition

At t = 0, the mass is at zero, and at rest, that is, $\mathrm{y}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0$ .

Now find the derivative of equation (2).

 

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{A}\left(-2{\mathrm{e}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+\varphi \right)+6{\mathrm{e}}^{-2\mathrm{t}}\cos \left(6\mathrm{t}+\varphi \right)\right)+\frac{4}{\sqrt{85}}\cos \left(2\mathrm{t}+\mathrm{\theta }\right)$

Now find the constants using the initial conditions.

 

 $$\begin{gathered} \mathrm{y}\left(0\right)={\mathrm{Ae}}^{-2\left(0\right)}\sin \left(6\left(0\right)+\varphi \right)+\frac{2}{\sqrt{85}}\sin \left(2\left(0\right)+\mathrm{\theta }\right) \\ 0=\mathrm{Asin}\left(\varphi \right)+\frac{2}{\sqrt{85}}\sin \left(\mathrm{\theta }\right) \end{gathered}$$

 So, $\mathrm{Asin}\left(\varphi \right)+\frac{2}{\sqrt{85}}\sin \left(\mathrm{\theta }\right)=0$ …… (3)

 $$\begin{gathered} \mathrm{y}\text{'}\left(0\right)=\mathrm{A}\left(-2{\mathrm{e}}^{-2\left(0\right)}\sin \left(6\left(0\right)+\varphi \right)+6{\mathrm{e}}^{-2\left(0\right)}\cos \left(6\left(0\right)+\varphi \right)\right)+\frac{4}{\sqrt{85}}\cos \left(2\left(0\right)+\mathrm{\theta }\right) \\ 0=-2\mathrm{Asin}\left(\varphi \right)+6\mathrm{Acos}\left(\varphi \right)+\frac{4}{\sqrt{85}}\cos \left(\mathrm{\theta }\right) \end{gathered}$$

 So, $-2\mathrm{Asin}\left(\varphi \right)+6\mathrm{Acos}\left(\varphi \right)+\frac{4}{\sqrt{85}}\cos \left(\mathrm{\theta }\right)=0$........(4)

Now solve the equation (3) and (4)

 

$$\begin{gathered} \frac{\mathrm{Asin}\left(\varphi \right)}{-2\mathrm{Asin}\left(\varphi \right)+6\mathrm{Acos}\left(\varphi \right)}=\frac{\frac{2}{\sqrt{85}}\sin \left(\mathrm{\theta }\right)}{\frac{4}{\sqrt{85}}\cos \left(\mathrm{\theta }\right)} \\ \frac{\tan \varphi }{2\tan \varphi -6}=\frac{1}{2}\mathrm{tan\theta } \\ \tan \varphi =\frac{9}{4}\left(2\tan \varphi -6\right) \\ =\frac{27}{11} \\ \varphi ={\tan }^{-1}\left(\frac{27}{11}\right)+\mathrm{k\pi } \end{gathered}$$

 

Then, find the value of A.

Using the fact of $\sin \left({\tan }^{-1}\left(\mathrm{x}\right)\right)=\frac{\mathrm{x}}{\sqrt{{\mathrm{x}}^2+1}}$ .

 $$\begin{gathered} \mathrm{A}=\frac{\frac{2}{\sqrt{85}}\sin \left({\tan }^{-1}\left(\frac{9}{2}\right)\right)}{-\sin \left({\tan }^{-1}\left(\frac{27}{11}\right)\right)} \\ =-\frac{2\sqrt{34}}{51} \end{gathered}$$

 

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=-\frac{2\sqrt{34}}{51}{\mathrm{e}}^{-2\mathrm{t}}\sin \left(6\mathrm{t}+{\tan }^{-1}\left(\frac{27}{11}\right)\right)+\frac{2}{\sqrt{85}}\sin \left(2\mathrm{t}+{\tan }^{-1}\left(\frac{9}{2}\right)\right)$  .

Then, the frequency of the motion is 

 

$$\begin{gathered} \frac{{\gamma }_{\mathrm{r}}}{2\mathrm{\pi }}=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{k}}{\mathrm{m}}-\frac{{\mathrm{b}}^2}{2{\mathrm{m}}^2}} \\ =\frac{1}{2\mathrm{\pi }}\sqrt{40-8} \\ =\frac{4\sqrt{2}}{2\mathrm{\pi }} \\ =\frac{2\sqrt{2}}{\mathrm{\pi }} \end{gathered}$$

So, the frequency is $\frac{2\sqrt{2}}{\mathrm{\pi }}$ .