Question

A force of 2 pounds stretches a spring $\mathbf{\text{1foot}}$. A mass weighing $\mathbf{\text{3.2pounds}}$is attached to the spring, and the system is then immersed in a medium that offers a damping force that is numerically equal to $\mathbf{0}\mathbf{.}\mathbf{4}$times the instantaneous velocity.

(a) Find the equation of motion if the mass is initially released from rest from a point 1 foot $\mathbf{\text{1foot}}$above the equilibrium position

(b) Express the equation of motion in the form given in $\mathbf{\left(}\mathbf{23}\mathbf{\right)}$.

(c) Find the fi­rst time at which the mass passes through the equilibrium position heading upward.

Short answer

(a) The equation of motion if the mass is initially released from rest from is $x\left(t\right)=e^{2t}(-\cos 4t-\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 4t)$.

(b) The equation of motion is $x\left(t\right)=\frac{\sqrt{5}}{2}{e}^{2t}\sin (4t+4.249)$.

(c) The mass passes through the equilibrium position heading upward for the first time when $t=1.294\text{s}$.

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:

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

Using Differential equation to find Amplitude and phase angle:

You have a system where an object with a mass of $m=\text{3.2lb}$attached to a spring with a constant $k$is exhibiting free damped motion where $\beta =0.4$. You know that the spring is stretched for $s=\text{1ft}$by a force $F=\text{2lb}$.

Your main objective is to find the equation of motion given the initial conditions and then write it in its alternative form. You also need to solve for the time when the mass passes through the equilibrium position while heading upwards for the first time.

Remember that to solve for the spring constant $k$, you make use of Hooke's law which has the formula:

$F=ks$ ….. (1)

Where, $F$is the force in pounds and $s$is the elongation in feet.

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

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

Where,${\omega }^2=\raisebox{1ex}{$k$}\!\left/ \!\raisebox{-1ex}{$m$}\right.$and$2\lambda =\raisebox{1ex}{$\beta $}\!\left/ \!\raisebox{-1ex}{$m$}\right.$.

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 overdamped 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:

$x\left(t\right)=e^{\lambda t}(c_1+c_{2}t)$ ..... (4)

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

$x\left(t\right)=e^{\lambda t}\left(c_1\cos \sqrt{{\lambda }^2-{\omega }^2}t+c_2\sin \sqrt{{\lambda }^2-{\omega }^2}t\right)$ ..... (5)

After finding the equation of motion, you can write it in its alternative form as,

$x\left(t\right)=A{e}^{\lambda t}\sin \left(\sqrt{{\omega }^2-{\lambda }^2}t+\varphi \right)$ ..... (6)

Where, the amplitude $A$is given by the formula,

$A=\sqrt{c_1^2+c_2^2}$

And the phase angle is solved using the formula,

$\tan \varphi =\frac{c_1}{c_2}$

(a) Find the equation of motion:

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

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{3.2\text{pounds}}{32{\displaystyle }{\displaystyle \raisebox{1ex}{$ft$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}\\ & =& \frac{1}{10}\text{slug}\end{array}$

Following Hooke's Law, we find the spring constant to be,

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

With an attached mass of $m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\text{slug},\beta =0.4$, and $k=2\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.$the differential equation modeling the free damped motion of the spring and mass system using 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}+\frac{0.4}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}\frac{dx}{dt}+\left(\frac{2}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}}\right)x& =& 0\end{array}$

$\frac{d^{2}x}{d{t}^2}+4\frac{dx}{dt}+20x=0$ ..... (7)

Following Eq. (2) and (7), you have

$\begin{array}{rcl}2\lambda & =& 4\\ \lambda & =& 2\end{array}$

And

$$\begin{gathered} {\omega }^2=20 \\ \omega =2\sqrt{5} \end{gathered}$$

It follows that,

$\begin{array}{rcl}{\lambda }^2-{\omega }^2& =& {\left(2\right)}^2-{\left(2\sqrt{5}\right)}^2\\ & =& -16\end{array}$

And since ${\lambda }^2-{\omega }^2<0$, you have an underdamped system whose solution is given by Eq. (3) as,

$\begin{array}{rcl}x\left(t\right)& =& e^{\lambda t}\left(c_1{e}^{\sqrt{{\lambda }^2{\omega }^2}t}+c_2{e}^{\sqrt{{\lambda }^2{\omega }^{2}t}}\right)\\ & =& e^{\lambda t}\left(c_1\cos \sqrt{{\omega }^2-{\lambda }^2}t+c_2\sin \sqrt{{\omega }^2-{\lambda }^2}t\right)\\ & =& e^{2t}\left(c_1\cos \sqrt{20-4}t+c_2\sin \sqrt{20-4}t\right)\end{array}$

$x\left(t\right)=e^{2t}\left(c_1\cos 4t+c_2\sin 4t\right)$ ..... (8)

The given also states the initial displacement to be at the equilibrium position So, $x\left(0\right)=-1$. You also have the initial velocity giving $x^{\text{'}}\left(0\right)=0$. Plugging to Eq. (8), you find,

$\begin{array}{rcl}1& =& e^0(c_1\cos 0+c_2\sin 0°)\\ -1& =& c_1\left(1\right)+c_2\left(0\right)\\ c_1& =& -1\end{array}$

From the derivative of Eq. (8) plugging $x^{\text{'}}\left(0\right)=0$,

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& e^{2t}\left(-4c_1\sin 4t+4c_2\cos 4t\right)-2e^{2t}\left(c_1\cos 4t+c_2\sin 4t\right)\\ 0& =& e^0\left(4\sin 0+4c_2\cos 0°\right)-2e^0\left(-\cos 0+c_2\sin 0°\right)\\ 0& =& 4c_2+2\\ c_2& =& -\frac{1}{2}\end{array}$

With $c_1=-1$and $c_2=-\frac{1}{2}$, the required equation of motion using Eq. (8) is,

$x\left(t\right)=e^{2t}\left(-\cos 4t-\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 4t\right)$

(b) Find the alternative equation of motion:

Representing this in terms of Eq. (4), you plug $c_1=-1$and $c_2=-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$to Eq. (5) giving the amplitude to be,

$\begin{array}{rcl}A& =& \sqrt{{\left(-1\right)}^2+{\left(-\frac{1}{2}\right)}^2}\\ & =& \frac{\sqrt{5}}{2}\end{array}$

Likewise, you plug the two values to Eq. (6) giving the phase angle as,

$\begin{array}{rcl}\varphi & =& {\tan }^1\left(\frac{-1}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\\ & =& 4.249\end{array}$

Thus, using Eq. (4) gives the alternative form of the equation of motion as,

$\begin{array}{rcl}x\left(t\right)& =& \frac{\sqrt{5}}{2}{e}^{2t}\sin \left(\sqrt{{\left(2\sqrt{5}\right)}^2-2^2}t+4.249\right)\\ & =& \frac{\sqrt{5}}{2}{e}^{2t}\sin (4t+4.249)\end{array}$

(c) Find the value of t:

To find the value of $t$where the mass passes through the equilibrium position, you set the equation of motion obtained given by Eq. (8) equal to zero. Solving for $\mathit{t}$, you get,

$0=\frac{\sqrt{5}}{2}{e}^{2t}\sin (4t+4.249)$

$0=e^{2t}\sin (4t+4.249)$ ..... (9)

The exponential function $e^{2t}$will never be zero in finite time, so the only time that $x\left(t\right)=0$is when,

$\sin (4t+4.249)=0$

Getting the inverse sine of both sides of Eq. (9), you find $t$to be,

$\begin{array}{rcl}{\sin }^1\left[\sin (4t+4.249)\right]& =& {\sin }^1\left(0\right)\\ 4t+4.249& =& 0+\pi n\\ 4t& =& -4.249+\pi n\end{array}$

$t=-1.062+\frac{\pi n}{4}$ ..... (10)

Where, $n$is any integer $(n=0,1,2,\dots )$.

The strategy is to find values of $n$that will satisfy the required conditions.

When $n=0$, Eq. (10) gives

$t=-1.062$

Which will disregard since $\mathit{t}$must be greater than zero for the motion to start.

When $n=1,t=-0.277$so you also disregard this value of $\mathit{t}$since it is below zero.

Then, plugging $n=2$to Eq. (10), you get $t=0.509$which is a possible value of $\mathit{t}$. But because the mass was initially released above the equilibrium position, $x\left(0\right)=-1$, it will be heading downwards as it passes through the equilibrium position.

Finally, when $n=3$the mass will now be heading upward as it passes through the equilibrium position. Using Eq. (10), you find,

$\begin{array}{rcl}t& =& -1.062+\frac{\pi n}{4}\\ & =& -1.062+\frac{3\pi }{4}\\ & =& 1.294\text{s}\end{array}$

Hence, the mass passes through the equilibrium position heading upward for the first time when$t=1.294\text{s}$.