Question

A mass weighing $\mathbf{\text{16pounds}}$stretches a spring $\raisebox{1ex}{$\mathbf{8}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{3}$}\right.\mathbf{\text{feet}}$. The mass is initially released from rest from a point $\mathbf{\text{2feet}}$below the equilibrium position, and the subsequent motion takes place in a medium that offers a damping force that is numerically equal to $\raisebox{1ex}{$\mathbf{1}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.$the instantaneous velocity. Find the equation of motion if the mass is driven by an external force equal to $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{10}\mathbf{cos}\mathbf{3}\mathit{t}$.

Short answer

The equation of motion is$x\left(t\right)=e^{-t/2}\left(-\frac{{\displaystyle 4}}{{\displaystyle 3}}\cos \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t-\frac{{\displaystyle 64}}{{\displaystyle 3\sqrt{47}}}\sin \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t\right)+\frac{10}{3}(\cos 3t+\sin 3t)$.

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:

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

Find the solution of the differential equation:

Converting the given weight to an engineering unit of mass gives,

$\begin{array}{rcl}m& =& \frac{W}{g}\\ & =& \frac{16\text{pounds}}{32{\displaystyle \raisebox{1ex}{$\text{ft}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.}}\\ & =& \frac{1}{2}\text{slug}\end{array}$

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

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

$$\begin{gathered} 16=k\left(\frac{8}{3}\right) \\ k=6\raisebox{1ex}{$\text{lb}$}\!\left/ \!\raisebox{-1ex}{$\text{ft}$}\right. \end{gathered}$$

You also have the damping constant as $\beta =\frac{1}{2}$ and the driving force $f\left(t\right)=10\cos 3t$.

From the values in Step l, the differential equation modelling the driven motion with damping of the spring and mass system is,

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right) \\ \frac{1}{2}\frac{d^{2}x}{d{t}^2}+\frac{1}{2}\frac{dx}{dt}+6x=10\cos 3t \\ \frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+12x=20\cos 3t \end{gathered}$$

The solution to the differential equation is given by,

$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$

Where the complimentary function $x_c\left(t\right)$is the solution to the homogenous differential equation,

$\frac{d^{2}x}{d{t}^2}+\frac{dx}{dt}+12x=0$ ….. (1)

And the particular function is the solution to be,

$x_p^{\text{'}\text{'}}+x_p^{\text{'}}+12x_p=20\cos 3t$ ….. (2)

Find the auxiliary equation and solve:

Using undetermined coefficients.

Solving for $x_c\left(t\right)$, you have the auxiliary equation of (1) to be,

$m^2+m+12=0$

This gives,

$m_{1,2}=-\frac{1}{2}\pm \frac{\sqrt{47}}{2}i$

Thus, you have

$x_c\left(t\right)=e^{-t/2}\left(c_1\cos \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t+c_2\sin \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t\right)$

Using undetermined coefficients, you have $x_p$and its derivatives to be,

$$\begin{gathered} x_p=A\cos 3t+B\sin 3t \\ {x}_p^{\text{'}}=-3A\sin 3t+3B\cos 3t \\ {x^{\text{'}\text{'}}}_p=-9A\cos 3t-9B\sin 3t \end{gathered}$$

Following equation (2) in Step 3, you have

$$\begin{gathered} -9A\cos 3t-9B\sin 3t-3A\sin 3t+3B\cos 3t+12(A\cos 3t+B\sin 3t)=20\cos 3t \\ 3A\cos 3t+3B\sin 3t-3A\sin 3t+3B\cos 3t=20\cos 3t \\ (3A+3B)\cos 3t+(-3A+3B)\sin 3t=20\cos 3t \end{gathered}$$

Equating the coefficients of the resulting equation in Step 5 gives,

$3A+3B=20$

And

$-3A+3B=0$

Solving this system of equations simultaneously yields $A=\frac{10}{3}$and $B=\frac{10}{3}$. Thus, the particular solution is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& \frac{10}{3}\cos 3t+\frac{10}{3}\sin 3t\\ & =& \frac{10}{3}(\cos 3t+\sin 3t)\end{array}$

Now that you have solved for $x_c\left(t\right)$and $x_p\left(t\right)$, it follows that,

$x\left(t\right)=e^{-t/2}(c_1\cos \frac{\sqrt{47}}{2}t+c_2\sin \frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos 3t+\sin 3t)$

However, there are still unknowns. You need to utilize the initial conditions to be able to find $c_1$and $c_2$.

You are given with $x\left(0\right)=2$and $x^{\text{'}}\left(0\right)=0$. Solving for $x\left(t\right)$, you get

$\begin{array}{rcl}2& =& e^0(c_1\cos 0+c_2\sin 0)+\frac{10}{3}(\cos 0°+\sin 0°)\\ & =& c_1+\frac{10}{3}\end{array}$

$c_1=-\frac{4}{3}$

Find the value of c:

Using the first derivative of $x\left(t\right)$to find $c_2$gives,

$\begin{array}{rcl}{x}^{\text{'}}\left(t\right)& =& e^{t/2}\left(-c_1\frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}\sin \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t+c_2\frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}\cos \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t\right)\\ & & -\frac{1}{2}{e}^{t/2}\left(c_1\cos \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t+c_2\sin \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t\right)+\frac{10}{3}\left(-3\sin 3t+3\cos 3t\right)\end{array}$

$\begin{array}{rcl}0& =& e^0\left(0+\frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}{c}_2\right)-\frac{1}{2}{e}^0(c_1+0)+10(0+1)\\ & =& \frac{\sqrt{47}}{2}{c}_2+\frac{2}{3}+10\end{array}$

$c_2=-\frac{64}{3\sqrt{47}}$

With $c_1=-\frac{4}{3}$and $c_2=-\frac{64}{3\sqrt{47}}$, you have the equation of motion to be,

$x\left(t\right)=e^{-t/2}\left(-\frac{{\displaystyle 4}}{{\displaystyle 3}}\cos \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t-\frac{{\displaystyle 64}}{{\displaystyle 3\sqrt{47}}}\sin \frac{{\displaystyle \sqrt{47}}}{{\displaystyle 2}}t\right)+\frac{10}{3}(\cos 3t+\sin 3t)$

Hence, the above equation is the required equation of motion.