Question

A$\mathbf{4}$-Pound weight stretches a spring$\mathbf{2}$feet. the weight is released from rest$\mathbf{18}$inches above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to$\frac{\mathbf{7}}{\mathbf{8}}$times the instantaneous velocity. use the Laplace transform to find the equation of motion x(t).

Short answer

Newton’s 2ndLaw

Newton's second law describes the effect of net force and mass on an object's acceleration.

$\begin{array}{l}ma=\sum F\\ ma=W-F_S-F_{\beta }\\ m\frac{d^{2}X}{d{t}^2}=mg-k\left(s+x\right)-\beta \frac{dX}{dt}\\ m\frac{d^{2}X}{d{t}^2}=mg-ks-kx-\beta \frac{dX}{dt}\end{array}$

$$\begin{gathered} X\left(s\right)=-\frac{3}{2}.\frac{s}{s^2+7s+16}-\frac{21}{2}.\frac{1}{s^2+7s+16} \\ x\left(t\right)=-\frac{3}{2}\cos \left(\sqrt{\frac{15}{2}}\right)e^{\frac{-7t}{2}}--\frac{7\sqrt{15}}{10}\sin \left(\frac{\sqrt{15}}{2}t\right)e^{\frac{-7t}{2}} \end{gathered}$$

Step-by-step solution

definition of Newton’s 2nd Law

Newton's second law describes the effect of net force and mass on an object's acceleration.

Draw the diagram:

Newton’s 2ndLaw

$\begin{array}{l}ma=\sum F\\ ma=W-F_S-F_{\beta }\\ m\frac{d^{2}X}{d{t}^2}=mg-k\left(s+x\right)-\beta \frac{dX}{dt}\\ m\frac{d^{2}X}{d{t}^2}=mg-ks-kx-\beta \frac{dX}{dt}\end{array}$

We know that mg – k s = 0 which leaves us with

$\begin{array}{l}m\frac{d^{2}x}{d{t}^2}=-kx-\beta \frac{dx}{dt}\\ \frac{d^{2}x}{d{t}^2}=-\frac{k}{m}.x-\frac{\beta }{m}.\frac{dx}{dt}\\ \frac{d^{2}x}{d{t}^2}+\frac{\beta }{m}.\frac{dx}{dt}+\frac{k}{m}.x=0\end{array}$

Plug inour values:

$\begin{array}{l}m=\frac{1}{8},\beta =\frac{7}{8},k=2\\ \frac{d^{2}x}{d{t}^2}+\left(\frac{\frac{7}{8}}{\frac{1}{8}}\right)\frac{dx}{dt}+\left(\frac{2}{\frac{1}{8}}\right)x=0\\ \frac{d^{2}x}{d{t}^2}+7\frac{dx}{dt}++16x=0\\ x\text{'}\text{'}+7x\text{'}+16x=0\end{array}$

Take the Laplace of both sides and apply initial conditions:

$\begin{array}{l}y\left(0\right)=\frac{-3}{2},y\text{'}\left(0\right)=0\\ L\left\{X\text{'}\text{'}\right\}=s^{2}X\left(s\right)-sx\left(0\right)-x\text{'}\left(0\right)\\ L\left\{x\text{'}\text{'}\right\}=s^{2}X\left(s\right)+s\left(\frac{3}{2}\right)\\ L\left\{x\text{'}\right\}=sX\left(s\right)-x\left(0\right)\\ L\left\{x\text{'}\right\}=sX\left(s\right)+\frac{3}{2}\\ L\left\{x\right\}=X\left(s\right)\\ L\left\{0\right\}=0\end{array}$

Step 2: Apply inverse Laplace transform

A transformation of a function is particularly useful in reducing the solution of an ordinary linear differential equation with constant coefficients to the solution of a polynomial equation.

Substitute into the differential equation

$\begin{array}{rcl}\left[s^{2}X\left(s\right)+s\left(\frac{3}{2}\right)\right]+7\left[sX\left(s\right)+s\left(\frac{3}{2}\right)\right]+16\left[X\left(s\right)\right]& =& 0\\ s^{2}X\left(s\right)+s\left(\frac{3}{2}\right)+7sX\left(s\right)+\frac{21}{2}+16X\left(s\right)& =& 0\\ s^{2}X\left(s\right)+7sX\left(s\right)+16X\left(s\right)& =& -s\left(\frac{3}{2}\right)-\frac{21}{2}\\ & & +-\\ & & +-\\ & & +-\\ X\left(s\right)\left[s^2+7s+16\right]& =& -s\left(\frac{3}{2}\right)-\frac{21}{2}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s}{s^2+7s+16}-\frac{21}{2}.\frac{1}{s^2+7s+16}\\ & & \end{array}$

Complete square:

$\begin{array}{rcl}X\left(s\right)& =& -\frac{3}{2}.\frac{s}{\left[s^2+7s+{\left(\frac{7}{2}\right)}^2\right]+16-{\left(\frac{7}{2}\right)}^2}-\frac{21}{2}.\frac{1}{\left[s^2+7s+{\left(\frac{7}{2}\right)}^2\right]+16-\left(7\right)}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}-\frac{21}{2}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s+\frac{7}{2}-\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}-\frac{21}{2}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s+\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}+\frac{3}{2}.\frac{\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}-\frac{21}{2}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s+\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}+\frac{21}{4}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}.\frac{21}{2}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\\ X\left(s\right)& =& -\frac{3}{2}.\frac{s+\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}-\frac{21}{4}.\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{7}{2}}\\ & & \end{array}$

Take the inverse transform of both sides

$\begin{array}{rcl}{L}^{-1}\left\{X\left(s\right)\right\}& =& -\frac{3}{2}{L}^{-1}\left\{\frac{s+\frac{7}{2}}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\right\}-\frac{21}{4}{L}^{-1}\left\{\frac{1}{{\left(s+\frac{7}{2}\right)}^2+\frac{15}{4}}\right\}\\ X\left(t\right)& =& -\frac{3}{2}{L}^{-1}\left\{\frac{s}{s^2+\frac{15}{4}}\right\}.e^{-\frac{7t}{2}-}\frac{21}{4}{L}^{-1}\left\{\frac{1}{s^2+\frac{15}{4}}\right\}.e^{-\frac{7t}{2}}\\ X\left(t\right)& =& -\frac{3}{2}{L}^{-1}\left\{\frac{s}{s^2+\frac{15}{4}}\right\}.e^{-\frac{7t}{2}-}\frac{21}{4}{L}^{-1}\left\{\frac{1}{s^2+\frac{15}{4}}\right\}.e^{-\frac{7t}{2}}\\ & & \end{array}$

$\begin{array}{rcl}X\left(t\right)& =& -\frac{3}{2}{L}^{-1}\left\{\frac{s}{s^2+\frac{15}{4}}\right\}.e^{-\frac{7t}{2}-}\frac{21}{4}{L}^{-1}\left\{\frac{1}{s^2+{\left(\sqrt{\frac{15}{2}}\right)}^2}\right\}.e^{-\frac{7t}{2}}\\ X\left(t\right)& =& -\frac{3}{2}{L}^{-1}\left\{\frac{s}{s^2+\frac{15}{2}}\right\}.e^{-\frac{7t}{2}-}.\frac{21}{4}.\frac{21}{\sqrt{15}}{L}^{-1}\left\{\frac{\sqrt{\frac{15}{2}}}{s^2+{\left(\sqrt{\frac{15}{2}}\right)}^2}\right\}.e^{-\frac{7t}{2}}\\ x\left(t\right)& =& -\frac{3}{2}\cos \left(\sqrt{\frac{15}{2}}\right)e^{-\frac{7t}{2}}--\frac{7\sqrt{15}}{10}\sin \left(\frac{\sqrt{15}}{2}t\right)e^{-\frac{7t}{2}}\end{array}$

Hence,

$\begin{array}{rcl}x\left(t\right)& =& -\frac{3}{2}\cos \left(\sqrt{\frac{15}{2}}\right)e^{-\frac{7t}{2}}--\frac{7\sqrt{15}}{10}\sin \left(\frac{\sqrt{15}}{2}t\right)e^{-\frac{7t}{2}}\end{array}$