Question

In parts (a) and (b) of Problem 27 determine whether the mass passes through the equilibrium position. In each case ­and the time at which the mass attains its extreme displacement from the equilibrium position. What is the position of the mass at this instant?

Short answer

The negative sign implies that the extreme displacement is above the equilibrium position.

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:

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

Equation of motion:

Referring the part (a) of the problem , you have the equation of motion as,

$x\left(t\right)=\frac{4}{3}{e}^{-2t}-\frac{1}{3}{e}^{-8t}$

To determine whether the mass passes through the equilibrium position, let $x\left(t\right)=0$then verify if you can find possible values of $t$.

Solving forlocalid="1668410111043" $t$whenlocalid="1668410096041" $x\left(t\right)=0$, you have

localid="1668410126388" $\begin{array}{rcl}0& =& \frac{4}{3}{e}^{-2t}-\frac{1}{3}{e}^{-8t}\\ e^{-8t}& =& 4e^{2t}\\ \frac{e^{-8t}}{e^{2t}}& =& 4\\ \ln e^{-6t}& =& \ln 4\end{array}$

localid="1668410141214" $\begin{array}{rcl}-6t& =& \ln 4\\ t& =& -0.2310\text{s}\end{array}$

Because localid="1668410152980" $\mathit{t}$must be greater than zero, it follows that the mass does not pass through the equilibrium position.

The mass attains extreme displacement when localid="1668410162874" $x^{\text{'}}\left(t\right)=0$. Getting the first derivative of (1) and equating it to zero, you find

localid="1668410174751" $$\begin{gathered} x^{\text{'}}\left(t\right)=-\frac{8}{3}{e}^{2t}+\frac{8}{3}{e}^{-8t} \\ 0=-e^{2t}+e^{-8t} \\ {e}^{-2t}\text{=}{e}^{-8t} \end{gathered}$$

Thus, you can say that localid="1668410183537" $x^{\text{'}}\left(t\right)$is never equal to zero for localid="1668410193312" $t>0$.

Divide the logarithm to solve t:

However, you can still divide both sides by $e^{2t}$then take the natural logarithm of both sides yielding,

$\begin{array}{rcl}1& =& \frac{e^{-8t}}{e^{2t}}\\ 1& =& e^{-6t}\\ \ln 1& =& \ln e^{-6t}\\ 0& =& -6t\end{array}$

$t=0$

The only time that the mass reaches extreme displacement is when localid="1668410226609" $t=0$or when the mass was initially released. And when localid="1668410234579" $t=0$, you have the position of the mass to be

localid="1668410244303" $\begin{array}{rcl}x\left(0\right)& =& \frac{4}{3}-\frac{1}{3}\\ & =& 1\text{m}\end{array}$

Referring the part (b) of the problem, you have the equation of motion as,

localid="1668410261715" $x\left(t\right)=-\frac{2}{3}{e}^{-2t}+\frac{5}{3}{e}^{-8t}$

Substitute the initial conditions:

To determine whether the mass passes through the equilibrium position, let $x\left(t\right)=0$then verify if we can find possible values of $\mathit{t}$.

Solving for $t$when localid="1668410288596" $x\left(t\right)=0$, you have

localid="1668410299924" $\begin{array}{rcl}0& =& -\frac{2}{3}{e}^{2t}+\frac{5}{3}{e}^{8t}\\ 2e^{2t}& =& 5e^{8t}\\ \frac{2}{5}& =& \frac{e^{8t}}{e^{2t}}\end{array}$

localid="1668410312425" $\begin{array}{rcl}\ln \frac{2}{5}& =& \ln e^{6t}\\ -0.9163& =& -6t\\ t& =& 0.1527\text{s}\end{array}$

Becauselocalid="1668410323623" $\mathit{t}$is greater than zero, it is valid and the mass passes through the equilibrium position.

The mass attains extreme displacement when localid="1668410332624" $x^{\text{'}}\left(t\right)=0$. Getting the first derivative of (2) and equating it to zero, you find,

localid="1668410346394" $\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& \frac{4}{3}{e}^{2t}-\frac{40}{3}{e}^{8t}\\ 0& =& 4e^{2t}-40e^{-8t}\\ 40e^{-8t}& =& 4e^{2t}\\ \frac{e^{-8t}}{e^{2t}}& =& \frac{4}{40}\end{array}$

$e^{-6t}=\frac{1}{10}$

Taking the natural logarithm of both sides, you now have,

localid="1668410360292" $\begin{array}{rcl}\ln e^{-6t}& =& \ln \frac{1}{10}\\ -6t& =& -2.3026\\ t& =& 0.3838\text{s}\end{array}$

Thus, extreme displacement of the mass is attained whenlocalid="1668410381944" $t=0.3838\text{s}$.

Plugging the value of obtained to equation (2), you find the position of the mass as it attains extreme displacement at,

localid="1668410395547" $\begin{array}{rcl}x\left(0.3838\right)& =& -\frac{2}{3}{e}^{-2\left(0.3838\right)}+\frac{5}{3}{e}^{-8\left(0.3838\right)}\\ & =& -0.2321\text{m}\end{array}$

The negative sign implies that the extreme displacement is above the equilibrium position.