Question

Damped Motion Assume that the model for the spring/mass system in Problem 33 is replaced by$\mathit{m}{\mathit{x}}^{\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathit{x}}^{\mathbf{\backslash }\mathbf{c}\mathbf{e}\mathbf{n}\mathbf{t}}\mathbf{+}\mathit{k}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}$

In other words, the system is free but is subjected to damping numerically equal to 2 times the instantaneous velocity. With the same initial conditions and spring constant as in Problem 33 , investigate whether a mass m can be found that will pass through the equilibrium position at$\mathit{t}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$ second.

Short answer

The initial-value problem is $m_2=\frac{5-\sqrt{25-{\pi }^2}}{n^2{\pi }^2}$

Step-by-step solution

To Find the initial value problem

The initial-value problem is

$x^{\backslash \mathrm{cent}\backslash \mathrm{cent}}+\frac{2}{m}{x}^{\backslash \mathrm{cent}}+\frac{k}{m}=0,x\left(0\right)=0,x^{\backslash \mathrm{cent}}\left(0\right)=v_0.$With $k=10,$the auxiliary equation has roots $\frac{-1\pm \sqrt{1-10m}}{m}.$Consider the three cases:

Find the solution of the differential equation 

Case 1: If $m=\frac{1}{10},$the roots are 10 and 10 . Hence, the solution of the differential equation is

$x\left(t\right)=c_1{e}^{-10t}+c_{2}t{e}^{-10t}$The initial conditions imply$c_1=0$and $c_2=v_0$and so

$x\left(t\right)=v_{0}t{e}^{-10t}$The condition $x\left(1\right)=0$ implies $v_0{e}^{-10}=0$ which is impossible because $v_{?}0.$

Case 2 : If $m<\frac{1}{10},$then the roots are distinct real

$m_{1,2}-\frac{1}{m}\pm \frac{\sqrt{1-10m}}{m}$and the solution of the differential equation is

$x\left(t\right)=c_1{e}^{m_{1}t}+c_2{e}^{m_{2}t}$The initial conditions imply ${\mathit{c}}_{\mathbf{1}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}\mathbf{=}\mathbf{0}$and $m_1{c}_1+m_2{c}_2=v_0$and so $c_1=v_0/\left(m_1-\right.\left.m_2\right),c_2=-v_0/\left(m_1-m_2\right),$ and

$x\left(t\right)=\frac{v_0}{m_1-m_2}\left(e^{m_{1}t}-e^{m_{2}t}\right).$Again,$x\left(1\right)=0$is impossible because $v_{0}0.$

Case 3: If$m>\frac{1}{10},$then

$m_{1,2}-\frac{1}{m}\pm \frac{\sqrt{10m-1}}{m}i$

and the solution of the differential equation is

$x\left(t\right)=c_{1}e-t/mcos\frac{\sqrt{10m-1}}{m}t+c_{2}e-t/msin\frac{\sqrt{10m-1}}{m}t$The initial conditions imply $c_1=0$and $c_2=\frac{m{v}_0}{\sqrt{10m-1}}.$So that

$x\left(t\right)=\frac{m{v}_0}{\sqrt{10m-1}}e-t/msin\frac{\sqrt{10m-1}}{m}t$

The condition $x\left(1\right)=0$implies

$\frac{m{v}_0}{\sqrt{10m-1}}e-1/msin\frac{\sqrt{10m-1}}{m}=0$and so

$sin\frac{\sqrt{10m-1}}{m}=0$Hence, $\frac{\sqrt{10m-1}}{m}=n\pi$and so

$m=\frac{5\pm \sqrt{25-n^2{\pi }^2}}{n^2{\pi }^2}$But $m_n$are all real. Hence,$25-n^2{\pi }^{2}0$and so $n\mathfrak{T}5/\pi \gg 1.59$.

So that$n=1.$ So that

$m=\frac{5+\sqrt{25-{\pi }^2}}{n^2{\pi }^2}$

and

$m_2=\frac{5-\sqrt{25-{\pi }^2}}{n^2{\pi }^2}$

Final proof

$m_2=\frac{5-\sqrt{25-{\pi }^2}}{n^2{\pi }^2}$