Question

A mass weighing$\mathbf{\text{24pounds}}$stretches a spring$\mathbf{\text{4feet}}$. The subsequent motion takes place in medium that offers a damping force numerically equal to localid="1664048610111" $\mathit{\beta }$$\mathbf{(}\mathit{\beta }\mathbf{>}\mathbf{\text{0}}\mathbf{)}$times the instantaneous velocity. If the mass is initially released from the equilibrium position with an upward velocity of $\mathbf{\text{2}}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$, show that $\mathit{\beta }\mathbf{>}\mathbf{\text{3}}\sqrt{\mathbf{\text{2}}}$ the equation of motion is

localid="1664048854781" style="max-width: none; vertical-align: -15px;" $\mathit{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{-}\mathbf{3}}{\sqrt{{\mathbf{\beta }}^{\mathbf{2}}\mathbf{-}\mathbf{18}}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{\beta t}\mathbf{/}\mathbf{3}}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\frac{\mathbf{2}}{\mathbf{3}}\sqrt{{\mathbf{\beta }}^{\mathbf{2}}\mathbf{-}\mathbf{18}}\mathit{t}$
.

Short answer

The equation of motion is $x\left(t\right)=\frac{-3}{\sqrt{{\beta }^2-18}}{e}^{2\beta t/3}\sinh \frac{2}{3}\sqrt{{\beta }^2-18t}$.

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:

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

Using the Newton’s second law and Hook’s law:

Let $m$be the mass attached, $k$the spring constant and $\beta$the positive damping constant. The Newton's second law for the system is,

$m\frac{d^{2}x}{d{t}^2}=-kx-\beta \frac{dx}{dt}$

Where, $x\left(t\right)$is the displacement from the equilibrium position.

The mass is given by,

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{24\text{pounds}}{32{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\\ & =& \frac{3}{4}\text{slug}\end{array}$

According to Hook's law, you can evaluate the spring constant$k$as,

$\begin{array}{rcl}k& =& \frac{W}{s}\\ & =& \frac{24\text{lb}}{4\text{ft}}\\ & =& 6\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.\end{array}$

Also$\beta =0.4$

So that,

$\frac{d^{2}x}{d{t}^2}+\frac{4\beta }{3}\frac{dx}{dt}+8x=0$

The catachrestic equation is,

$m^2+\frac{4\beta }{3}m+8=0$

And also,

$m_{1,2}=-\frac{2}{3}\beta \pm \frac{2}{3}\sqrt{{\beta }^2-18}$

Find the general solution and substitution:

The general solution is,

$x\left(t\right)=e^{\frac{2}{3}\beta t}\left(c_1\cosh \frac{{\displaystyle 2}}{{\displaystyle 3}}\sqrt{{\beta }^2-18}t+c_2\sinh \frac{{\displaystyle 2}}{{\displaystyle 3}}\sqrt{{\beta }^2-18}t\right)$

According the initial conditions,

$x\left(0\right)=0,x^{\text{'}}\left(0\right)=2$

You get,

$c_1=0,,c_2=\frac{-3}{\sqrt{{\beta }^2-18}}$

And so,

$x\left(t\right)=\frac{-3}{\sqrt{{\beta }^2-18}}{e}^{2\beta t/3}\sinh \frac{2}{3}\sqrt{{\beta }^2-18t}$

Hence, it’s proved.