Question

When a mass of $\mathbf{\text{2kilograms}}$2 kilograms is attached to a spring whose constant is$\mathbf{\text{32}}\raisebox{1ex}{$\mathbf{\text{N}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{m}}$}\right.$, it comes to rest in the equilibrium position. Starting at$\mathbf{\text{t}}\mathbf{=}\mathbf{\text{0}}$, a force equal to$\mathbf{\text{f}}\mathbf{\left(}\mathbf{\text{t}}\mathbf{\right)}\mathbf{=}{\mathbf{\text{68e}}}^{\mathbf{\text{-2t}}}\mathbf{\text{cos4t}}$ is applied to the system. Find the equation of motion in the absence of damping.

Answer:

Short answer

The equation of motion in the absence of damping is,

$\begin{array}{rcl}x\left(t\right)& =& -\frac{1}{2}\cos 4t+\frac{9}{4}\sin 4t+e^{-2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t)\right)\end{array}$

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$m=\text{2kg},k=3\text{2}\raisebox{1ex}{$\text{N}$}\!\left/ \!\raisebox{-1ex}{$\text{m}$}\right.$, and$f\left(t\right)=68e^{2t}\cos 4t$, the differential equation modelling the driven motion without damping of the spring and mass system is,

$$\begin{gathered} m\frac{d^{2}x}{d{t}^2}+kx=f\left(t\right) \\ 2\frac{d^{2}x}{d{t}^2}+32x=68e^{2t}\cos 4t \end{gathered}$$

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

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

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

$2x_p^{\text{'}\text{'}}+32x_p=68e^{-2t}\cos 4t$

Solving for , you have the auxiliary equation as,

$m^2+16=0$

This gives,

$m_{1,2}=0\pm 4i$

Thus, the general solution to (1) is,

$x_c\left(t\right)=c_1\cos 4t+c_2\sin 4t$

Find the Complementary and particular function:

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

$$\begin{gathered} x_p=e^{-2t}(A\cos 4t+B\sin 4t) \\ {x}_p^{\text{'}}=e^{-2t}\left[(-4B-2A)\cos 4t+(-4A-2B)\sin 4t\right] \\ {x^{\text{'}\text{'}}}_p=e^{2t}\left[(-12A-16B)\cos 4t+(16A-12B)\sin 4t\right] \end{gathered}$$

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

$$\begin{gathered} 68e^{-2t}\cos 4t=2e^{-2t}\left[(-12A-16B)\cos 4t+(16A-12B)\sin 4t\right]+32e^{2t}(A\cos 4t+B\sin 4t) \\ 68e^{2t}\cos 4t=e^{2t}\left[(8A-32B)\cos 4t+(32A+8B)\sin 4t\right] \end{gathered}$$

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

$8A-32B=68$

And

$32A+8B=0$

Solving this system of equations simultaneously yields $A=\frac{1}{2}$and $B=-2$. Thus, the particular solution to (2) is,

$\begin{array}{rcl}{x}_p\left(t\right)& =& e^{-2t}(A\cos 4t+B\sin 4t)\\ & =& e^{-2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t\right)\end{array}$

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

$x\left(t\right)=c_1\cos 4t+c_2\sin 4t+e^{-2t}(\frac{1}{2}\cos 4t-2\sin 4t)$

Substitute the initial condition:

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

You are given with the initial displacement $x\left(0\right)=0$to solve for and get,

$\begin{array}{rcl}0& =& c_1\cos 0°+c_2\sin 0°+e^0\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 0°-2\sin 0°\right)\\ & =& c_1+0+\frac{1}{2}\end{array}$

$c_1=-\frac{1}{2}$

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

$$\begin{gathered} x^{\text{'}}\left(t\right)=-4c_1\sin 4t+4c_2\cos 4t+e^{2t}(-2\sin 4t-8\cos 4t)-2e^{2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t\right) \\ 0=2\sin 0°+4c_2\cos 0°+e^0(0-8)-2e^0\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}-0\right) \\ 0=4c_2-8-1 \\ {c}_2=\frac{9}{4} \end{gathered}$$

With $c_1=-\frac{1}{2}$and $c_2=\frac{9}{4}$, you have the equation of motion to be,

$\begin{array}{rcl}x\left(t\right)& =& c_1\cos 4t+c_2\sin 4t+e^{2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t\right)\\ & =& -\frac{1}{2}\cos 4t+\frac{9}{4}\sin 4t+e^{-2t}\left(\frac{{\displaystyle 1}}{{\displaystyle 2}}\cos 4t-2\sin 4t)\right)\end{array}$