Q42E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Forced Vibrations. As discussed in Section 4.1, a vibrating spring with damping that is under external force can be modeled by 

 $\mathit{m}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{b}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{k}\mathit{y}\mathbf{=}\mathit{g}\left(\mathit{t}\right)$, 

Where m > 0 is the mass of the spring system, b > 0 is the damping constant, k > 0 is the spring constant, g(t) is the force on the system at time t, and y(t) is the displacement from the equilibrium of the spring system at time t. Assume ${\mathit{b}}^{\mathbf{2}}\mathbf{<}\mathbf{4}\mathit{m}\mathit{k}$.

1. Determine the form of the equation of motion for the spring system when $\mathit{g}\left(\mathit{t}\right)\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{\beta }\mathit{t}$ by finding a general solution to equation (15). 
2. Discuss the long-term behavior of this system.

[Hint: Consider what happens to the general solution obtained in part (a) as $\mathit{t}\mathbf{\to }\mathbf{+}\mathbf{\infty }$.]

Step-by-Step Solution

Answer

1. The answer is:            $$\begin{gathered} y=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]sin\beta t \\ +\left[\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)}\right]cos\beta t \end{gathered}$$
2. As $t\to +\infty$ the system oscillates between $-\sqrt{A^2+B^2}$ to $\sqrt{A^2+B^2}$

Step 1: Firstly, write the given equation

Consider the differential equation as:

 $my\text{'}\text{'}+by\text{'}+ky=g\left(t\right); b^2<4mk ...\left(1\right)$

 

And 

$g\left(t\right)=sin\beta t$ 

Write the homogeneous differential equation of equation (1),

$my\text{'}\text{'}+by\text{'}+ky=0$

Step 2: Now find the complementary solution of the given equation is

The auxiliary equation for the above equation, $m{r}^2+br+k=0$

Solve the auxiliary equation,

$$\begin{gathered} m{r}^2+br+k=0 \\ r=\frac{-b\pm \sqrt{b^2-4mk}}{2m} \end{gathered}$$ 

 

One has,

$b^2-4mk<0$ 

 

The roots of the auxiliary equation are:

$m_1=\frac{-b+\sqrt{4mk-b^2}}{2m}, m_2=\frac{-b-\sqrt{4mk-b^2}}{2m}$

 

The complimentary solution of the given equation is,

$y_c=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)$

Step 3: find the particular solution to find a general solution for the equation

Assume, the particular solution of equation (1),

$y_p\left(t\right)=Asin\beta t+Bcos\beta t ...\left(2\right)$ 

 

Now find the first and second derivatives of the above equation,

$$\begin{gathered} y_p\text{'}\left(t\right)=A\beta cos\beta t-B\beta sin\beta t \\ {y}_p\text{'}\text{'}\left(t\right)=-A{\beta }^{2}sin\beta t-B{\beta }^{2}cos\beta t \end{gathered}$$ 

 

Substitute the value of  $g\left(t\right), y_p\left(t\right), y_p\text{'}\left(t\right)$ and $y_p\text{'}\text{'}\left(t\right)$ in the equation (1),

 $$\begin{gathered} my\text{'}\text{'}+by\text{'}+ky=g\left(t\right) \\ m\left[-A{\beta }^{2}sin\beta t-B{\beta }^{2}cos\beta t\right]+b\left[A\beta cos\beta t-B\beta sin\beta t\right]+k\left[Asin\beta t+Bcos\beta t\right]=sin\beta t \\ \left[kA-mA{\beta }^2-bB\beta \right]sin\beta t+\left[kB-mB{\beta }^2+bA\beta \right]cos\beta t=sin\beta t \end{gathered}$$

 

Compare the coefficient of the above equation,

$$\begin{gathered} kA-mA{\beta }^2-bB\beta =1 ...\left(3\right) \\ kB-mB{\beta }^2+bA\beta =0 ...\left(4\right) \end{gathered}$$ 

From equation (4),

$$\begin{gathered} kB-mB{\beta }^2+bA\beta =0 \\ B\left(k-m{\beta }^2\right)=-bA\beta \\ B=A\frac{-b\beta }{k-m{\beta }^2} ...\left(5\right) \end{gathered}$$

Substitute the value of B in the equation (3),

$$\begin{gathered} kA-mA{\beta }^2-bB\beta =1 \\ kA-mA{\beta }^2-b\left[A\frac{-b\beta }{k-m{\beta }^2}\right]\beta =1 \\ A\left[k-m{\beta }^2+\frac{{\left(b\beta \right)}^2}{k-m{\beta }^2}\right]=1 \\ A\left[\frac{k\left(k-m{\beta }^2\right)-m{\beta }^2\left(k-m{\beta }^2\right)+{\left(b\beta \right)}^2}{k-m{\beta }^2}\right]=1 \\ A\left[\frac{k^2-2km{\beta }^2+{\left(m{\beta }^2\right)}^2+{\left(b\beta \right)}^2}{k-m{\beta }^2}\right]=1 \\ A=\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2} \end{gathered}$$ 

 

Substitute the value of A in the equation (5),

$$\begin{gathered} B=A\frac{-b\beta }{k-m{\beta }^2} \\ B=\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]\frac{-b\beta }{k-m{\beta }^2} \\ B=\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)} \end{gathered}$$ 

 

Substitute the value of A and B in equation (2).

 

Therefore, the particular solution of equation (1),

$y_p\left(t\right)=\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]sin\beta t+\left[\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)}\right]cos\beta t$

Step 4: Now, Find the general solution,

Therefore, the general solution is,

$$\begin{gathered} y=y_c\left(t\right)+y_p\left(t\right) \\ y=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]sin\beta t \\ +\left[\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)}\right]cos\beta t ...\left(3\right) \end{gathered}$$

Step 5: Discuss the long-term behavior of this system at t &#8594; + &#8734;

Given, as $t\to +\infty$

$$\begin{gathered} y=c_1{e}^{-\frac{bt}{2m}}cos\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+c_2{e}^{-\frac{bt}{2m}}sin\left(\frac{\sqrt{4km-b^2}}{2m}t\right)+\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]sin\beta t \\ +\left[\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)}\right]cos\beta t \\ As t\to \infty \\ y=+\left[\frac{k-m{\beta }^2}{{\left(m{\beta }^2-k\right)}^2+{\left(b\beta \right)}^2}\right]sin\beta t+\left[\frac{-b\beta }{{\left(m{\beta }^2-k\right)}^2+\left(b\beta \right)}\right]cos\beta t \end{gathered}$$ 

 

In this function, one can write,

$$\begin{gathered} Asin\beta t+Bcos\beta t=\sqrt{A^2+B^2}\left[\frac{A}{\sqrt{A^2+B^2}}sin\beta t+\frac{B}{\sqrt{A^2+B^2}}cos\beta t\right] \\ =\sqrt{A^2+B^2}sin\left(\beta t+\alpha \right) \end{gathered}$$ 

 

Where, 

$\alpha =ta{n}^{-1}\left(\frac{B}{A}\right)$ 

The range of $sin\left(\beta t+\alpha \right)$ is [-1, -1]

Therefore, as $t\to +\infty$ the system oscillates between $-\sqrt{A^2+B^2}$to $\sqrt{A^2+B^2}$.