Question

A $\mathbf{\text{4foot}}$spring measures $\mathbf{\text{8feet}}$long after a mass weighing $\mathbf{\text{8pounds}}$is attached to it. The medium through which the mass moves offers a damping force numerically equal to $\sqrt{\mathbf{\text{2}}}$times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of $\mathbf{\text{5}}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$. Find 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 mass has a maximum displacement of $0.6503\text{ft}$at $\text{0.35355}\text{s}$.

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:

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

Damped motion:

You have a system where an object weighing $W=\text{8lb}$attached to a spring with a constant $k$, is being stretched from $\text{4ft}$to $\text{8ft}$. You also know that the object was initially released from the equilibrium position having a downward velocity of $\text{5}\raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{$\text{s}$}\right.$and that the medium has a damping coefficient of $\beta =\sqrt{2}$.

Your main objective is to find the equation of motion given the initial conditions, then you will use it to solve for our secondary objective which is to find the values of variables related to its extreme displacements such as the time and position.

To solve for $k$, you can use Hooke's law whose formula is:

$F=ks$ ..... (1)

Where, $k$is the constant of proportionality and $s$is the elongation in feet.

Because there is damping, the system exhibits free damped motion described by the differential equation,

$\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0$ ..... (2)

Here,

${\omega }^2=\raisebox{1ex}{$k$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

And

$2\lambda =\raisebox{1ex}{$\beta $}\!\left/ \!\raisebox{-1ex}{$m$}\right.$

Solve the second order differential equation:

Note that this is a second-order homogenous linear differential equation with constant coefficients having three cases for the solution.

When ${\lambda }^2-{\omega }^2>0$, the system is over damped and the solution is:

$x\left(t\right)=e^{\lambda t}\left(c_1{e}^{\sqrt{{\lambda }^2{\omega }^{2}t}}+c_2{e}^{\sqrt{{\lambda }^2{\omega }^{2}t}}\right)$

Then when ${\lambda }^2-{\omega }^2=0$, the system is critically damped and the solution is given by,

$x\left(t\right)=e^{\lambda t}(c_1+c_{2}t)$

Lastly, when ${\lambda }^2-{\omega }^2<0$, the system is underdamped having the solution to be,

$x\left(t\right)=e^{\lambda t}\left(c_1\cos \sqrt{{\lambda }^2-{\omega }^2}t+c_2\sin \sqrt{{\lambda }^2-{\omega }^2}t\right)$

Following Eq. (1), you find the spring constant to be

$\begin{array}{rcl}F& =& W\\ & =& ks\\ 8& =& k(8-4)\\ k& =& 2\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.\end{array}$

Converting the given weight to an appropriate unit of mass gives you,

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{8\text{pounds}}{32{\displaystyle \raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}\\ & =& \frac{1}{4}\text{slug}\end{array}$

Since the initial displacement is at the equilibrium position, it implies that $x\left(0\right)=0$. The initial velocity gives you $x^{\text{'}}\left(0\right)=5$. In addition, we have the damping constant to be $\beta =\sqrt{2}$.

With $m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$slug, $\beta =\sqrt{2}$, and $k=2\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.$, the differential equation modeling the free damped motion of the spring-mass system following Eq. (2) is,

$\begin{array}{rcl}\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x& =& 0\\ \frac{d^{2}x}{d{t}^2}+\left(\frac{\sqrt{2}}{1/4}\right)\frac{dx}{dt}+\left(\frac{2}{1/4}\right)x& =& 0\end{array}$

$\frac{d^{2}x}{d{t}^2}+4\sqrt{2}\frac{dx}{dt}+8x=0$ ..... (3)

Since,

$\begin{array}{rcl}{\lambda }^2-{\omega }^2& =& \left(2{\sqrt{2}}^2\right)-{\left(2\sqrt{2}\right)}^2\\ & =& 0\end{array}$

Solve the equation to find motion:

You have a critically damped system whose solution is given by Eq. (3.2) as:

$\begin{array}{rcl}x\left(t\right)& =& e^{\lambda t}(c_1+c_{2}t)\end{array}$

$\begin{array}{rcl}x\left(t\right)& =& e^{2\sqrt{2}t}(c_1+c_{2}t)\end{array}$ ..... (4)

Using the initial condition$x\left(0\right)=0$on Eq. (4), we find

$$\begin{gathered} 0=e^0(c_1+c_2\cdot 0) \\ {c}_1=0 \end{gathered}$$

Getting the first derivative of Eq. (4), plugging $x^{\text{'}}\left(0\right)=5$
leads to:

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& e^{2\sqrt{2}t}\left(0+c_2\right)+\left(c_1+c_{2}t\right)\left(-2\sqrt{2}{e}^{2\sqrt{2}t}\right)\\ 5& =& c_2{e}^{2\sqrt{2}\cdot 0}-(c_2\cdot 0)2\sqrt{2}{e}^{2\sqrt{2}\cdot 0}\\ c_2& =& 5\end{array}$

Thus, the required equation of motion is,

$x\left(t\right)=5t{e}^{2\sqrt{2}t}$ ..... (5)

Find the value of t:

The mass reaches extreme displacement at $x^{\text{'}}\left(t\right)=0$.

Taking the first derivative of Eq. (5) using the product rule then solving for $t$, you obtain,

$\begin{array}{rcl}x\left(t\right)& =& 5t{e}^{2\sqrt{2}t}\\ x^{\text{'}}\left(t\right)& =& 5\left(-2\sqrt{2}{e}^{2\sqrt{2}t}t+e^{2\sqrt{2}t}\right)\\ \frac{0}{5}& =& -2\sqrt{2}{e}^{2\sqrt{2}t}t+e^{2\sqrt{2}t}\\ 2\sqrt{2}{e}^{2\sqrt{2}t}t& =& e^{2\sqrt{2}t}\end{array}$

$\begin{array}{rcl}t& =& \frac{e^{2\sqrt{2}t}}{2\sqrt{2}{e}^{2\sqrt{2}t}}\\ & =& \frac{1}{2\sqrt{2}}\\ & =& \frac{\sqrt{2}}{4}\text{s}\\ & =& 0.35355\text{s}\end{array}$

Substituting the previously obtained value of $t=\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$to the equation of motion given by Eq. (5), you find the position when the mass is at the extreme displacement to be

$\begin{array}{rcl}x\left(\frac{\sqrt{2}}{4}\right)& =& 5\left(\frac{{\displaystyle \sqrt{2}}}{{\displaystyle 4}}\right)e^{2\sqrt{2}\cdot \frac{\sqrt{2}}{4}}\\ & =& \frac{5\sqrt{2}}{4}\cdot e^{-1}\\ & =& \frac{5\sqrt{2}}{4e}\text{ft}\\ & =& 0.6503\text{ft}\end{array}$

Hence, the mass has a maximum displacement of $0.6503\text{ft}$at $\text{0.35355}\text{s}$.