Q4 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.

Determine the equation of motion for an undamped system at resonance governed by

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{y}=5\mathrm{cost} \\ \mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1 \end{gathered}$$

Sketch the solution.

Step-by-Step Solution

Answer

Therefore, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\sin \left(\mathrm{t}\right)+\frac{5}{2}\mathrm{tsin}\left(\mathrm{t}\right)$ and its sketch is shown below.

Step 1: General form

The angular frequency:

 

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

 

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

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 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$ .

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, 

 $$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{y}=5\mathrm{cost}; \\ \mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1. \end{gathered}$$

Then, m = 1, k = 1, and  ${\mathrm{F}}_0=5 and \gamma =1$.

 Find the $\omega$ value.

 $$\begin{gathered} \backslash b\omega =\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \\ =\sqrt{\frac{1}{1}} \\ =1 \end{gathered}$$

.

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

Find the derivative of y.

$\mathrm{y}\text{'}\left(\mathrm{t}\right)=\mathrm{A}\omega \cos \left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \mathrm{t}+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tcos}\omega \mathrm{t}$

Step 3: Implement the initial conditions.

Given the initial conditions are $\mathrm{y}\left(0\right)=0,\mathrm{y}\text{'}\left(0\right)=1$.

 Then,

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ \mathrm{y}\left(0\right)=\mathrm{Asin}\left(\omega \left(0\right)+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\left(0\right)\sin \omega \left(0\right) \\ 0=\mathrm{Asin}\left(\varphi \right) \end{gathered}$$

And

$$\begin{gathered} \left(\mathrm{t}\right)=\mathrm{A}\omega \cos \left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \mathrm{t}+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tcos}\omega \mathrm{t} \\ \mathrm{y}\text{'}\left(0\right)=\mathrm{A}\omega \cos \left(\omega \left(0\right)+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\sin \omega \left(0\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\left(0\right)\cos \omega \left(0\right) \\ 0=\mathrm{A}\omega \cos \left(\varphi \right) \\ 1=\mathrm{Acos}\left(\varphi \right) \end{gathered}$$

So, A cannot be zero because $0=\mathrm{Asin}\left(\varphi \right)$.

 

Since $\sin \left(\varphi \right)=0$. Then, $\varphi ={\sin }^{-1}\left(0\right)=\mathrm{k\pi }$, Where k belongs to an integer.

Step 4: Find the solution.

Case (1):

 

If k is even, k = 2l. then A becomes 1 and the solution can be written as

$$\begin{gathered} \left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}+\mathrm{k\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}+2\mathrm{l\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \end{gathered}$$

 

Case (2):

 

If k is odd, k = 2l + 1, then A becomes -1 and the solution can be written as;

$$\begin{gathered} \left(\mathrm{t}\right)=-\sin \left(\omega \mathrm{t}+2\mathrm{l\pi }+\mathrm{\pi }\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \end{gathered}$$

Since both cases are shown $\mathrm{y}\left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$. Then,

$$\begin{gathered} \left(\mathrm{t}\right)=\sin \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t} \\ =\sin \left(\mathrm{t}\right)+\frac{5}{2\times 1\times 1}\mathrm{tsin}\left(\mathrm{t}\right) \\ =\sin \left(\mathrm{t}\right)+\frac{5}{2}\mathrm{tsin}\left(\mathrm{t}\right) \end{gathered}$$

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\sin \left(\mathrm{t}\right)+\frac{5}{2}\mathrm{tsin}\left(\mathrm{t}\right)$.

A sketch of the solution is shown below.