Question

After a mass weighing $\mathbf{\text{10pounds}}$is attached to a 5-foot spring, the spring measures $\mathbf{\text{7feet}}$. This mass is removed and replaced with another mass that weighs $\mathbf{\text{8pounds}}$. The entire system is placed in a medium that offers a damping force that is numerically equal to the instantaneous velocity.

(a) Find the equation of motion if the mass is initially released from a point $\raisebox{1ex}{$\mathbf{\text{1}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{2}}$}\right.\mathbf{\text{foot}}$below the equilibrium position with a downward velocity of $\mathbf{\text{1}}\raisebox{1ex}{$\mathbf{\text{ft}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{s}}$}\right.$.

(b) Express the equation of motion in the form given in (23).

(c) Find the times at which the mass passes through the equilibrium position heading downward.

(d) Graph the equation of motion.

Short answer

(a) The equation of motion is $x\left(t\right)=e^{2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t+\frac{{\displaystyle 1}}{{\displaystyle 2}}\sin 4t\right)$.

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

(c) The time at which the mass passes through the equilibrium position heading downward is $t=\frac{(8n-1)\pi }{16}\text{s},n=1,2,3,\dots$.

(d) The graph of the equation of motion is,

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:

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

Homogeneous differential equation:

You have a system where an object with a mass of $m=\text{8lb}$attached to a spring with a constant $k$is exhibiting free damped motion where $\beta =1$. You also know that the spring is stretched for $\text{5ft}$to $\text{7ft}$by a force $F=\text{10lb}$.

Your main objective is to find the equation of motion given the initial conditions $x\left(0\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$and $x^{\text{'}}\left(0\right)=1$; then write it in its alternative form. You also need to solve for the times when the mass passes through the equilibrium position while heading downwards.

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,

$\frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0$ ..... (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 over damped and the solution is,

$x\left(t\right)=e^{\lambda t}\left(c_1{e}^{\sqrt{{\lambda }^2{\omega }^2}t}+c_2{e}^{\sqrt{{\lambda }^2{\omega }^2}t}\right)$ ….. (3)

Find the amplitude and phase angle:

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 mass and substitute the initial condition:

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

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

Following Hooke's Law given by Eq. (1), you find the spring constant to be,

$\begin{array}{rcl}F& =& W\\ & =& ks\end{array}$

$$\begin{gathered} 10=k(7-5) \\ k=5\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right. \end{gathered}$$

With an attached mass of $m=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\text{slug},\beta =1,$and $k=5\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right.$the differential equation modelling the free damped motion of the spring and mass system using Eq. (2) is,

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}+2\lambda \frac{dx}{dt}+{\omega }^{2}x=0 \\ \frac{d^{2}x}{d{t}^2}+\left(\frac{1}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\right)\frac{dx}{dt}+\left(\frac{{\displaystyle 5}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}\right)x=0 \end{gathered}$$

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

$\begin{array}{rcl}{\lambda }^2-{\omega }^2& =& {\left(2\right)}^2-{\left(2\sqrt{5}\right)}^2\\ & =& -6\end{array}$

And since , you have an underdamped system whose solution is given by Eq. (3).

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

Plugging the initial condition $x\left(0\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$to Eq. (8), you can find,

$$\begin{gathered} \frac{1}{2}=e^0(c_1\cos 0°+c_2\sin 0°) \\ \frac{1}{2}=c_1\left(1\right)+c_2\left(0\right) \\ {c}_1=\frac{1}{2} \end{gathered}$$

Likewise, if you plug $x^{\text{'}}\left(0\right)=1$to the derivative of Eq. (8) you get,

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

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

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

(b) Substitution:

Representing this in terms of Eq. (4), you need to substitute$c_1=\frac{1}{2}$and $c_2=\frac{1}{2}$to Eq. (5), which leads to,

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

Note that $c_1>0$and $c_2>0$, so must lie in the first quadrant. With Eq. (5), it follows that,

$\begin{array}{rcl}\varphi & =& {\tan }^1\left(\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right)\\ & =& \frac{\pi }{4}\end{array}$

Therefore, you obtain the alternative form of the equation of motion using Eq. (4) as,

$\begin{array}{rcl}x\left(t\right)& =& A{e}^{-\lambda t}\sin \left(\sqrt{{\omega }^2-{\lambda }^2}t+\varphi \right)\\ & =& \frac{\sqrt{2}}{2}{e}^{2t}\sin \left(4t+\frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)\end{array}$

(c) Find the value of t:

To find the values of $t$when the mass passes through the equilibrium position, you set the equation of motion given by Eq. (9) to be equal to zero. Solving for $t$, you get,

$\begin{array}{rcl}0& =& \frac{\sqrt{2}}{2}{e}^{2t}\sin \left(4t+\frac{\pi }{4}\right)\\ & =& e^{2t}\sin \left(4t+\frac{\pi }{4}\right)\\ & =& \sin \left(4t+\frac{\pi }{4}\right)\end{array}$

The exponential function $e^{2t}$will not be zero in finite time, so the only time that $x\left(t\right)=0$is when,

$\sin \left(4t+\frac{\pi }{4}\right)=0$

Getting the inverse sine of both sides of Eq. (8), you find $t$to be,

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

$t=-\frac{\pi }{16}+\frac{\pi n}{4}$

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

When $n=0$, Eq. (9) gives $t=-\frac{\pi }{16}$which you will disregard since must be greater than zero for the motion to start.

Then, when $n=1$is plugged to Eq. (9) you get $t=\frac{3\pi }{16}$. You now have a possible value of $t$. But because the mass was initially released below the equilibrium position, $x\left(0\right)=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$, it will be heading upward as it passes through the equilibrium position.

Finally, when $n=2$Eq. (9) gives $t=\raisebox{1ex}{$7\pi $}\!\left/ \!\raisebox{-1ex}{$16$}\right.$which is a positive value. So, the mass will now be heading downwards as it passes through the equilibrium position. So, $n$must be a multiple of $2$. You now rewrite Eq. (9) as,

$\begin{array}{rcl}t& =& -\frac{\pi }{16}+\frac{\pi \left(2n\right)}{4}\\ & =& -\frac{\pi }{16}+\frac{8\pi n}{16}\\ & =& \frac{(8n-1)\pi }{16}\text{s}\end{array}$

Where,$n\ge 1(n=1,2,3,\dots )$

(d) Graph the equation of motion:

Using a graphing utility, the graph of the equation of motion is shown in the figure below.