Question

Can there be beats when a damping force is added to the model in part (a) of Problem 43? Defend your position with graphs obtained either from the explicit solution of the problem $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{2}\mathit{\lambda }\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}{\mathit{F}}_{\mathbf{0}}\mathit{c}\mathit{o}\mathit{s}\mathit{\gamma }\mathit{t}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$, or from solution curves obtained using a numerical solver.

Short answer

The final answer is,

$\begin{array}{rcl}x\left(t\right)& =& e^{-\lambda t}\left(\frac{{\displaystyle F_0({\omega }^2-{\lambda }^2)}}{{\displaystyle {({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2}}\cos \sqrt{({\omega }^2-{\lambda }^2)}t-\frac{{\displaystyle 2y{F}_0}}{{\displaystyle {({\omega }^2-{\lambda }^2)}^2-4{\gamma }^2{\lambda }^2\sqrt{({\omega }^2-{\lambda }^2)}}}\sin \sqrt{({\omega }^2-{\lambda }^2)t}\right)\\ & & +\frac{2\lambda \gamma F_0}{({\omega }^2-{\lambda }^2)-4{\gamma }^2{\lambda }^2}\sin \gamma t\end{array}$

Step-by-step solution

Definition:

A physical system consisting of a body B passing through a medium M. In order for B to move, it needs to push M out of the way to make room for it. Hence M exerts a force upon B which opposes the motion of B. This force opposing the motion of B is referred to as a damping force.

Solving the homogeneous differential equation:

Solve the homogeneous differential equation as below.

$x\text{'}\text{'}+2\lambda x\text{'}+{\omega }^{2}x=0$

From the characteristic equation,

$m^2+2\lambda m+{\omega }^2=0$

You find

$\begin{array}{rcl}{m}_1& =& -\lambda -\sqrt{{\lambda }^2-{\omega }^2}\\ & =& -\lambda -i\sqrt{{\omega }^2-{\lambda }^2}\end{array}$

And

$\begin{array}{rcl}{m}_2& =& -\lambda +\sqrt{{\lambda }^2-{\omega }^2}\\ & =& -\lambda +i\sqrt{{\omega }^2-{\lambda }^2}\end{array}$

Therefore,

$\begin{array}{rcl}{x}_c& =& c_1{e}^{(-\lambda -i\sqrt{{\omega }^2-{\lambda }^2})t}+c_2{e}^{(-\lambda +i\sqrt{{\omega }^2-{\lambda }^2})t}\\ & =& e^{-\lambda t}\left[c_1{e}^{\left(i\sqrt{{\omega }^2-{\lambda }^2}\right)t}+c_2{e}^{\left(i\sqrt{{\omega }^2-{\lambda }^2}\right)t}\right]\\ & =& e^{-\lambda t}\left[\left(c_1\cos \sqrt{{\omega }^2-{\lambda }^2}t\right)-c_1\sin \sqrt{{\omega }^2-{\lambda }^2}t+c_2\sin \sqrt{{\omega }^2-{\lambda }^2}t\right]\\ & =& e^{-\lambda t}\left(c_{3}cos\sqrt{{\omega }^2-{\lambda }^2}t+c_4\sqrt{{\omega }^2-{\lambda }^2}t\right)\end{array}$

Where,

$$\begin{gathered} c_3=c_1+c_2 \\ {c}_4=i(c_2-c_1) \end{gathered}$$

Finding the general solution:

Define the general solution as below.

$x\left(t\right)=e^{-\lambda t}(c_{3}cos\sqrt{{\omega }^2-{\lambda }^2}t+c_4\sqrt{{\omega }^2-{\lambda }^2}t)+\frac{F_0({\omega }^2-{\lambda }^2)}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2}\cos \gamma t+\frac{2\lambda \gamma F_0}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2}\sin yt$

$$\begin{gathered} c_3=-\frac{F_0({\omega }^2-{\lambda }^2)}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2} \\ x\text{'}\left(0\right)=0 \\ 0=-\lambda c_4\sqrt{{\omega }^2-{\lambda }^2}+\frac{2\gamma \lambda F_0}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2} \\ {c}_4=-\frac{2\gamma F_0}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2\sqrt{{\omega }^2-{\lambda }^2}}\sin \sqrt{{\omega }^2-{\lambda }^2}t \end{gathered}$$

The graph equation $x\left(t\right)$with $\omega =\sqrt{2}.\gamma =1,\lambda =1,F_0=1$.

Beats of t>0:

Draw the graph below.

Hence the final answer is,

$\begin{array}{rcl}x\left(t\right)& =& e^{-\lambda t}\left(\frac{F_0({\omega }^2-{\lambda }^2)}{{({\omega }^2-{\lambda }^2)}^2-4{\lambda }^2{\gamma }^2}\cos \sqrt{({\omega }^2-{\lambda }^2)}t-\frac{2y{F}_0}{{({\omega }^2-{\lambda }^2)}^2-4{\gamma }^2{\lambda }^2\sqrt{({\omega }^2-{\lambda }^2)}}\sin \sqrt{({\omega }^2-{\lambda }^2)t}\right)\\ & & +\frac{2\lambda \gamma F_0}{({\omega }^2-{\lambda }^2)-4{\gamma }^2{\lambda }^2}\sin \gamma t\end{array}$