Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Undamped oscillators that are driven at resonance have unusual (and nonphysical) solutions.

1. To investigate this, find the synchronous solution $\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{\Omega }\mathit{t}\mathbf{+}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\Omega }\mathit{t}$ to the generic forced oscillator equation $\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathit{m}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{b}\mathit{y}\text{'}\mathbf{+}\mathit{k}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{\Omega }\mathit{t}$.
2. Sketch graphs of the coefficients A and B, as functions of $\mathit{\Omega }$ for $\mathit{m}\mathbf{=}\mathbf{1}\mathbf{,}\mathit{b}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathit{k}\mathbf{=}\mathbf{25}$.
3. Now set $\mathit{b}\mathbf{=}\mathbf{0}$ in your formulas for A and B and re-sketch the graphs in part (b), with $\mathit{m}\mathbf{=}\mathbf{1}$, and $\mathit{k}\mathbf{=}\mathbf{25}$. What happens at $\mathit{\Omega }\mathbf{=}\mathbf{5}$? Notice that the amplitudes of the synchronous solutions grow without bound as $\mathit{\Omega }$ approaches 5.
4. Show directly, by substituting the form $\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{\Omega }\mathit{t}\mathbf{+}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\Omega }\mathit{t}$ into equation (7), that when $\mathit{b}\mathbf{=}\mathbf{0}$ there are no synchronous solutions if $\mathit{\Omega }\mathbf{=}\sqrt{\mathit{k}\mathbf{/}\mathit{m}}$.
5. Verify that ${\mathbf{\left(}\mathbf{2}\mathit{m}\mathit{\Omega }\mathbf{\right)}}^{\mathbf{-}\mathbf{1}}\mathit{t}\mathit{s}\mathit{i}\mathit{n}\mathit{\Omega }\mathit{t}$ solves equation (7) when $\mathit{b}\mathbf{=}\mathbf{0}$ and $\mathit{\Omega }\mathbf{=}\sqrt{\mathit{k}\mathbf{/}\mathit{m}}$. 

Notice that this nonsynchronous solution grows in time, without bound.

Clearly one cannot neglect damping in analyzing an oscillator forced at resonance, because otherwise the solutions, as shown in part (e), are nonphysical. This behavior will be studied later in this chapter.

Step-by-Step Solution

Answer

a. The value of $A=\frac{k-m{\Omega }^2}{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$ and $B=\frac{b\Omega }{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$

b. The graph will be:

 

c. The re-sketched graph will be:

d. The system does not have synchronous solutions when $b=0$ and $\Omega =\sqrt{\frac{k}{m}}$ then $k=25$.

e. The given $y=(2m\Omega {)}^{-1}tsin\Omega t$ is a non-synchronous solution and it grows in time without any bound.

Step 1: Differentiate the given equation

Given differential equation is $my\text{'}\text{'}+by\text{'}+ky=cos\Omega t$

 

Let $y=Acos\Omega t+Bsin\Omega t$ then

 $$\begin{gathered} y\text{'}=-A\Omega sin\Omega t+B\Omega sin\Omega t \\ y\text{'}\text{'}=-A{\Omega }^{2}cos\Omega t-B{\Omega }^{2}sin\Omega t \end{gathered}$$

Step 2: Substitute the values

Substitute these in the differential equations and find the solution:

$$\begin{gathered} my\text{'}\text{'}+by\text{'}+ky=-m{\Omega }^2(Acos\Omega t+Bsin\Omega t)+b\Omega (-Asin\Omega t+Bcos\Omega t)+k(Acos\Omega t+Bsin\Omega t) \\ \left(-Am{\Omega }^2+Bb\Omega +Ak\right)cos\Omega t+\left(-Bm{\Omega }^2-Ab\Omega +Bk\right)sin\Omega =cos\Omega t \end{gathered}$$

  

Compare the coefficients on both sides:

 $$\begin{gathered} -Am{\Omega }^2+Bb\Omega +Ak=1 \\ A\left(k-m{\Omega }^2\right)+Bb\Omega =1 \\ B\left(k-m{\Omega }^2\right)-Ab\Omega =0 \end{gathered}$$

Step 3: Finding the A, B

By solving these two equations one can get

 $A=\frac{k-m{\Omega }^2}{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$ and $B=\frac{b\Omega }{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$.

 

Now $b=0.1$ with $m=1$ and $k=25$ then the graph is shown below blue color gives A graph and blue color is B graph. And here A and B is taken along $\mathrm{y}-\mathrm{axis}$ and $\Omega$ is taken along $x-axis$

Step 4: Finding the A, B

If $b=0,m=1,k=25$ Then,

$A=\frac{25-{\Omega }^2}{{\left(25-{\Omega }^2\right)}^2+0\times {\Omega }^2}\Rightarrow \frac{1}{25-{\Omega }^2}$

And

 $B=\frac{0\times \Omega }{{\left(25-{\Omega }^2\right)}^2+0^2\times {\Omega }^2}\Rightarrow B=0$

  

And if $\Omega =5$ then the solution does not exist for this system

Step 5: Check the equation is synchronous or not

From part (a) we have A and  B

$A=\frac{k-m{\Omega }^2}{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$ 

 

And

$B=\frac{b\Omega }{{\left(k-m{\Omega }^2\right)}^2+b^2{\Omega }^2}$ 

 

And if $b=0$ and $\Omega =\sqrt{\frac{k}{m}}$ then $k=25$. So, the system does not have synchronous solutions.

Step 6: Substitute the values in differential equation

Let $(y=(2m\Omega {)}^{-1}tsin\Omega t)$ then

$$\begin{gathered} y\text{'}=(2m\Omega {)}^{-1}(sin\Omega t+\Omega tcos\Omega t) \\ y\text{'}\text{'}=(2m\Omega {)}^{-1}\left(2\Omega cos\Omega t-{\Omega }^{2}tsin\Omega t\right) \end{gathered}$$ 

  

Substitute these in the differential equation

 $my\text{'}\text{'}+ky=m\times (2m\Omega {)}^{-1}\left(2\Omega cos\Omega t-{\Omega }^{2}tsin\Omega t\right)+k\times (2m\Omega {)}^{-1}tsin\Omega t$

 

Substitute $\Omega =\sqrt{\frac{k}{m}}$ 

                    $=(2\Omega {)}^{-1}\left(2\Omega cos\Omega t-{\Omega }^{2}tsin\Omega t\right)+(2\Omega {)}^{-1}tsin\Omega t=cos\Omega t$

Therefore $y=(2m\Omega {)}^{-1}tsin\Omega t$ is a non-synchronous solution and it grows in time without any bound.