Question

A mass of $\mathbf{\text{1slug}}$, when attached to a spring, stretches it $\mathbf{\text{2feet}}$and then comes to rest in the equilibrium position. Starting at $\mathit{t}\mathbf{=}\mathbf{0}$, an external force equal to $\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{8}\mathbf{sin}\mathbf{4}\mathit{t}$ is applied to the system. Find the equation of motion if the surrounding medium offers a damping force that is numerically equal to $\mathbf{8}$times the instantaneous velocity.

Short answer

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

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:

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

$m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right)$

$m\frac{d^{2}x}{d{t}^2}+\beta \frac{dx}{dt}+kx=f\left(t\right)$ ….. (1)

The solution of the resulting differential equation in Step 2 is given by,

$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$ ….. (2)

Where the complementary 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,

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

Solve the auxiliary equation:

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

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

Which 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,

$$\begin{gathered} x_p=A\cos 4t+B\sin 4t \\ {x}_p^{\text{'}}=-4A\sin 4t+4B\cos 4t \\ {x^{\text{'}\text{'}}}_p=-16A\cos 4t-16B\sin 4t \end{gathered}$$

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

$\begin{array}{rcl}8\sin 4t& =& -16A\cos 4t-16B\sin 4t+8(-4A\sin 4t+4B\cos 4t)+16(A\cos 4t+B\sin 4t)\\ & =& -32A\sin 4t+32B\cos 4t\end{array}$

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

$-32A+0B=8$

And

$0A+32B=0$

Solving this system of equations simultaneously yields $A=\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$and $B=0$. Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& A\cos 4t+B\sin 4t\\ & =& -\frac{1}{4}\cos 4t\end{array}$

Particular solution and substitution:

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

$x\left(t\right)=c_1{e}^{-4t}+c_{2}t{e}^{-4t}-\frac{1}{4}\cos 4t$

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 $x\left(0\right)=0$you solve for $c_1$and get

$\begin{array}{rcl}0& =& c_1{e}^0+c_2\left(0\right)e^0-\frac{1}{4}\cos 0°\\ & =& c_1-\frac{1}{4}\end{array}$

$c_1=\frac{1}{4}$

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

$$\begin{gathered} x^{\text{'}}\left(t\right)=-4c_1{e}^{4t}+c_2(-4t{e}^{4t}+e^{4t})+\sin 4t \\ 0=-e^0+c_2(0+e^0)+\sin 0 \\ 0=-1+c_2+0 \\ {c}_2=1 \end{gathered}$$

With $c_1=\frac{1}{4}$and $c_2=1$, we have the equation of motion to be,

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