Question

In Problem 37 write the equation of motion in the form $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{\omega }\mathbf{t}\mathbf{+}\mathbf{\varphi }\mathbf{)}\mathbf{+}\mathit{B}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathit{s}\mathit{i}\mathit{n}\mathbf{(}\mathbf{4}\mathbf{t}\mathbf{+}\mathbf{\theta }\mathbf{)}$. What is the amplitude of vibrations after a very long time?

Short answer

The required equation of motion and the amplitude of vibrations after a very long time is$x\left(t\right)=\frac{\sqrt{85}}{4}\sin (4t-0.2187)+\frac{\sqrt{17}}{2}{e}^{-2t}\sin (4t-0.2450)$and$A=\frac{\sqrt{85}}{4}$respectively.

Step-by-step solution

Definition of Spring / Mass Systems:

Suppose you take into consideration an external force $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$acting on a vibrating mass on a spring. For example,$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$could represent a driving force causing an oscillatory vertical motion of the support of the spring. The inclusion of$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$in the formulation of Newton’s second law gives the differential equation of driven of forced motion:

$\mathit{m}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}\mathbf{-}\mathit{\beta }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}$

Undammed motion equation:

The equation of motion describing the undammed motion in the previous problem is given by,

$x\left(t\right)=-\frac{1}{2}\cos 4t+\frac{9}{4}\sin 4t_{x_c}+e^{2t}{\left(\frac{1}{2}\cos 4t-2\sin 4t\right)}_{x_p}$

Your goal is to be able to write the right side of the equation in the form as below.

$\begin{array}{rcl}{x}_c& =& -\frac{1}{2}\cos 4t+\frac{9}{4}\sin 4t\\ & =& A\sin (\omega t+\varphi )\end{array}$

And

$\begin{array}{rcl}{x}_p& =& e^{2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t\right)\\ & =& B{e}^{-2t}\sin (4t+\theta )\end{array}$

For equation (l) you have $c_1=-\frac{1}{2},c_2=\frac{9}{4}$and $\omega =4$. Solving for $A$, you can find,

$\begin{array}{rcl}A& =& \sqrt{{\left(-\frac{1}{2}\right)}^2+{\left(\frac{9}{4}\right)}^2}\\ & =& \frac{\sqrt{85}}{4}\end{array}$

Using Phase angle formula:

Then using the formula $\tan \varphi =\frac{c_1}{c_2}$, you get

$\begin{array}{rcl}\varphi & =& {\tan }^{-1}\left(\frac{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\right)\\ & =& -0.2187\end{array}$

Thus, you can now write as,

$x_c=\frac{\sqrt{85}}{4}\sin (4t-0.2187)$

For equation (2) you have $c_1=\frac{1}{2},c_2=-2$. Solving for $B$, you can find,

$\begin{array}{rcl}B& =& \sqrt{{\left(\frac{1}{2}\right)}^2+{(-2)}^2}\\ & =& \frac{\sqrt{17}}{2}\end{array}$

Then using the formula $\tan \varphi =\frac{c_1}{c_2}$, you get

$\begin{array}{rcl}\varphi & =& {\tan }^{-1}\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{-2}\right)\\ & =& -0.2450\end{array}$

Thus, you can now write $x_p$as,

$x_p=\frac{\sqrt{17}}{2}{e}^{-2t}\sin (4t-0.2450)$

The equation of motion and the graph:

Finally, you combine the results of Step 3 and Step 4 and obtain the required equation of motion as,

$x\left(t\right)=\frac{\sqrt{85}}{4}\sin (4t-0.2187)+\frac{\sqrt{17}}{2}{e}^{-2t}\sin (4t-0.2450)$

After a very long time or as $t\to \infty$, the exponential function $e^{-2t}$approaches zero. Mathematically, you have,

$\underset{t\to \infty }{\lim }x\left(t\right)=\frac{\sqrt{85}}{4}\sin (4t-0.2187)$

Thus, the amplitude approaches $A=\frac{\sqrt{85}}{4}$as $t\to \infty$
.

Hence, the required equation of motion is,

$x\left(t\right)=\frac{\sqrt{85}}{4}\sin (4t-0.2187)+\frac{\sqrt{17}}{2}{e}^{-2t}\sin (4t-0.2450)$

The graph:

Draw the required graph for equation of motion $x\left(t\right)$as below.