Question

In Problem 35 determine the equation of motion if the external force is $\mathit{f}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{-}\mathbf{t}}\mathit{s}\mathit{i}\mathit{n}\mathbf{4}\mathit{t}$. Analyze the displacements for $\mathit{t}\mathbf{\to }\mathit{\infty }$.

Short answer

The displacement for $t\to \infty$is,

$x\left(t\right)=\frac{1}{625}(24+100t)e^{-4t}-\frac{1}{625}{e}^{-t}(24\cos 4t+7\sin 4t);x\to 0$

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

Differential equation:

A mass of $\text{1slug}$exerts a force of $\text{32lb}$. Following Hooke's Law, you find the spring constant to be,

$$\begin{gathered} F=ks \\ 32=k\left(2\right) \end{gathered}$$

From the given values $m=1\text{slug},\beta =8,$and $f\left(t\right)=e^{-t}\sin 4t,$the differential equation modeling the driven motion with damping of the spring and mass system is

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right) \\ \frac{d^{2}x}{d{t}^2}+8\frac{dx}{dt}+16x=e^{-t}\sin 4t \end{gathered}$$

The solution of the resulting differential equation in Step 2 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.

$\frac{d^{2}x}{d{t}^2}+8\frac{dx}{dt}+16x=0$

And the particular function $x_p\left(t\right)$is the solution to be,

$x_p^{\text{'}\text{'}}+8x_p^{\text{'}}+16x_p=e^t\sin 4t$

Solve the complementary and particular function:

Solving for $x_c\left(t\right)$, you have the auxiliary equation as,

$m^2+8m+16=0$

Gives,

$m_1=m_2=-4$

Thus, the general solution to (1) is,

$x_c\left(t\right)=c_1{e}^{-4t}+c_{2}t{e}^{-4t}$

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

$$\begin{gathered} x_p=e^{-t}(A\cos 4t+B\sin 4t) \\ {x}_p^{\text{'}}=e^t[(-4B-A)\cos 4t-(4A+B)\sin 4t] \\ {x^{\text{'}\text{'}}}_p=e^{-t}\left[(-15A-8B)\cos 4t+(8A-15B)\sin 4t\right] \end{gathered}$$

Following equation (2) in Step 2, you have,

$\begin{array}{rcl}{e}^{-t}\sin 4t& =& e^{-t}[(-15A-8B)\cos 4t+(8A-15B)\sin 4t]+8e^{-t}[(-4B-A)\cos 4t-(4A+B)\sin 4t]\\ & & +16e^{-t}(A\cos 4t+B\sin 4t)\end{array}$

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

$-7A+24B=0$

And

$-24A-7B=1$

Solving this system of equations simultaneously yields $A=-\raisebox{1ex}{$24$}\!\left/ \!\raisebox{-1ex}{$625$}\right.$and $B=-\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$625$}\right.$. Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& e^t(A\cos 4t+B\sin 4t)\\ & =& e^{-t}\left(-\frac{{\displaystyle 24}}{{\displaystyle 625}}\cos 4t-\frac{{\displaystyle 7}}{{\displaystyle 625}}\sin 4t\right)\end{array}$

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

$x\left(t\right)=c_1{e}^{-4t}+c_{2}t{e}^{-4t}+e^{-t}\left(-\frac{{\displaystyle 24}}{{\displaystyle 625}}\cos 4t-\frac{{\displaystyle 7}}{{\displaystyle 625}}\sin 4t\right)$

Substitute the initial conditions and simplify:

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 the initial displacement $x\left(0\right)=0$so solve for $c_1$and get,

$$\begin{gathered} 0=c_1{e}^0+c_2\left(0\right)e^0+e^0\left(-\frac{{\displaystyle 24}}{{\displaystyle 625}}\cos 0°-\frac{{\displaystyle 7}}{{\displaystyle 625}}\sin 0°\right) \\ 0=c_1+0-\frac{24}{625}-0 \\ {c}_1=\frac{24}{625} \end{gathered}$$

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

$$\begin{gathered} x^{\text{'}}\left(t\right)=-4c_1{e}^{-4t}+c_2\left(-4t{e}^{-4t}+e^{4t}\right)+e^{-t}\left(\frac{{\displaystyle 96}}{{\displaystyle 625}}\sin 4t-\frac{{\displaystyle 28}}{{\displaystyle 625}}\cos 4t\right)-e^t\left(\frac{{\displaystyle 24}}{{\displaystyle 625}}\cos 0°-\frac{{\displaystyle 7}}{{\displaystyle 625}}\sin 0°\right) \\ 0=-\frac{96}{625}+c_2(0+1)+e^0\left(0-\frac{{\displaystyle 28}}{{\displaystyle 625}}\right)-e^0\left(-\frac{{\displaystyle 24}}{{\displaystyle 625}}-0\right) \\ {c}_2=\frac{96}{625}+\frac{28}{625}-\frac{24}{625} \\ {c}_2=\frac{100}{625} \end{gathered}$$

You can simplify $c_2=\frac{4}{25}$but you can simplify by factoring if $c_2=\frac{100}{625}$.

With $c_2=\frac{100}{625}$and $c_1=\frac{24}{625}$, you have the equation of motion to be,

$\begin{array}{rcl}x\left(t\right)& =& \frac{24}{625}{e}^{-4t}+\frac{100}{625}t{e}^{-4t}+e^{-t}\left(-\frac{{\displaystyle 24}}{{\displaystyle 625}}\cos 4t-\frac{{\displaystyle 7}}{{\displaystyle 625}}\sin 4t\right)\\ & =& \frac{1}{625}\left(24+100t\right)e^{-4t}-\frac{1}{625}{e}^{-t}(24\cos 4t+7\sin 4t)\end{array}$

As $t\to \infty$, the exponential functions $e^{-t}$and $e^{-4t}$both approach zero. Thus, we can say that the mass stops moving at large values of $t$or mathematically,

$\underset{t\to \infty }{\lim }x\left(t\right)=\frac{1}{625}(24+100t)\times 0-\frac{1}{625}\times 0(24\cos 4t+7\sin 4t)$

$x\left(t\right)=\frac{1}{625}(24+100t)e^{-4t}-\frac{1}{625}{e}^{-t}(24\cos 4t+7\sin 4t);x\to 0$