Q5 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.

An undamped system is governed by

$$\begin{gathered} \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}; \\ \mathrm{y}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0, \end{gathered}$$

Where, $\gamma \ne \omega :=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

1. Find the equation of motion of the system.
2. Use trigonometric identities to show that the solution can be written in the form data-custom-editor="chemistry" $\mathrm{y}\left(\mathrm{t}\right)=\frac{2{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\sin \left(\frac{\omega +\gamma }{2}\mathrm{t}\right)\sin \left(\frac{\omega -\gamma }{2}\mathrm{t}\right)$.
3. When is near, then is small, while is relatively large compared with. Hence, y(t) can be viewed as the product of a slowly varying sine function and a rapidly varying sine function. The net effect is a sine function y(t) with frequency, which serves as the time-varying amplitude of a sine function with frequency. This vibration phenomenon is referred to as beats and is used in turning stringed instruments. This same phenomenon in electronics is called amplitude modulation. To illustrate this phenomenon, sketch the curve y(t) for ${\mathrm{F}}_0=32,\mathrm{m}=2,\omega =9$ and $\gamma =7$ .

Step-by-Step Solution

Answer

a. Therefore, the equation of motion of the system is $\mathrm{y}\left(\mathrm{t}\right)=-\frac{{\mathrm{F}}_0}{\mathrm{k}-\mathrm{m}{\gamma }^2}\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{{\mathrm{F}}_0\cos \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2}$ .

 

b. Hence, the given statement is true. That is, the solution can be written in the form of $\mathrm{y}\left(\mathrm{t}\right)=\frac{2{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\sin \left(\frac{\omega +\gamma }{2}\mathrm{t}\right)\sin \left(\frac{\omega -\gamma }{2}\mathrm{t}\right)$.

c. Accordingly, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\frac{2}{5}\sin \left(5\mathrm{t}\right)\sin \left(4\mathrm{t}\right)$ 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 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 data-custom-editor="chemistry" $\mathrm{y}\left(\mathrm{t}\right)=\mathrm{Asin}\left(\omega \mathrm{t}+\varphi \right)+\frac{{\mathrm{F}}_0}{2\mathrm{m}\omega }\mathrm{tsin}\omega \mathrm{t}$.

Step 2: Evaluate the equation

Given that,

 $$\begin{gathered} \mathrm{m}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dt}}^2}+\mathrm{ky}={\mathrm{F}}_0\cos \gamma \mathrm{t}; \\ \mathrm{y}\left(0\right)=\mathrm{y}\text{'}\left(0\right)=0, \end{gathered}$$

Then, the $\omega$value is $\omega =\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$.

Then, the general solution is $\mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{{\mathrm{F}}_0\cos \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2}$.

Find the derivative of y.

 $\mathrm{y}\text{'}\left(\mathrm{t}\right)=-\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_1\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_2\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{-\gamma {\mathrm{F}}_0\sin \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2}$

Step 3: Implement the initial conditions.

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

 Then,

 $$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{{\mathrm{F}}_0\cos \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2} \\ \mathrm{y}\left(0\right)={\mathrm{c}}_1\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\left(0\right)\right)+{\mathrm{c}}_2\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\left(0\right)\right)+\frac{{\mathrm{F}}_0\cos \gamma \left(0\right)}{\mathrm{k}-\mathrm{m}{\gamma }^2} \\ 0={\mathrm{c}}_1+\frac{{\mathrm{F}}_0}{\mathrm{k}-\mathrm{m}{\gamma }^2} \\ {\mathrm{c}}_1=-\frac{{\mathrm{F}}_0}{\mathrm{k}-\mathrm{m}{\gamma }^2} \end{gathered}$$

 

And

 

$$\begin{gathered} \mathrm{y}\text{'}\left(\mathrm{t}\right)=-\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_1\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_2\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{-\gamma {\mathrm{F}}_0\sin \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2} \\ \mathrm{y}\text{'}\left(0\right)=-\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_1\sin \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\left(0\right)\right)+\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_2\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\left(0\right)\right)+\frac{-\gamma {\mathrm{F}}_0\sin \gamma \left(0\right)}{\mathrm{k}-\mathrm{m}{\gamma }^2} \\ 0=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}{\mathrm{c}}_2 \\ {\mathrm{c}}_2=0 \end{gathered}$$

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=-\frac{{\mathrm{F}}_0}{\mathrm{k}-\mathrm{m}{\gamma }^2}\cos \left(\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}\mathrm{t}\right)+\frac{{\mathrm{F}}_0\cos \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2}$.

Step 4: Find the solution.

To prove: The solution can be written in the form;

$\mathrm{y}\left(\mathrm{t}\right)=\frac{2{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\sin \left(\frac{\omega +\gamma }{2}\mathrm{t}\right)\sin \left(\frac{\omega -\gamma }{2}\mathrm{t}\right)$

Since the trigonometric identity is $\mathrm{cosC}-\mathrm{cosD}=2\sin \left(\frac{\mathrm{C}+\mathrm{D}}{2}\right)\sin \left(\frac{\mathrm{C}-\mathrm{D}}{2}\right)$.

Then, rewrite the solution in the above form.

 $$\begin{gathered} -\frac{{\mathrm{F}}_0}{\mathrm{k}-\mathrm{m}{\gamma }^2}\cos \left(\omega \mathrm{t}\right)+\frac{{\mathrm{F}}_0\cos \gamma \mathrm{t}}{\mathrm{k}-\mathrm{m}{\gamma }^2}=\frac{{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\left(\cos \gamma \mathrm{t}-\cos \omega \mathrm{t}\right) \\ =\frac{2{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\sin \left(\frac{\omega +\gamma }{2}\mathrm{t}\right)\sin \left(\frac{\omega -\gamma }{2}\mathrm{t}\right) \end{gathered}$$

So, the solution can be rewritten in the form;

 $\mathrm{y}\left(\mathrm{t}\right)=\frac{2{\mathrm{F}}_0}{\mathrm{m}\left({\omega }^2-{\gamma }^2\right)}\sin \left(\frac{\omega +\gamma }{2}\mathrm{t}\right)\sin \left(\frac{\omega -\gamma }{2}\mathrm{t}\right)$

.

 Given conditions are ${\mathrm{F}}_0=32,\mathrm{m}=2,\omega =9$  and $\gamma =7$ .

 

Then substitute the values in the above equation.

 $$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)=\frac{2\times 32}{2\left(9^2-1^2\right)}\sin \left(\frac{9+1}{2}\mathrm{t}\right)\sin \left(\frac{9-1}{2}\mathrm{t}\right) \\ =\frac{32}{80}\sin \left(\frac{10}{2}\mathrm{t}\right)\sin \left(\frac{8}{2}\mathrm{t}\right) \\ =\frac{4}{10}\sin \left(5\mathrm{t}\right)\sin \left(4\mathrm{t}\right) \\ =\frac{2}{5}\sin \left(5\mathrm{t}\right)\sin \left(4\mathrm{t}\right) \end{gathered}$$

So, the solution is $\mathrm{y}\left(\mathrm{t}\right)=\frac{2}{5}\sin \left(5\mathrm{t}\right)\sin \left(4\mathrm{t}\right)$.

A sketch of the solution is shown below.