Chapter 4, Problem 4.64P

To determine

The equation of motion for the system.

Answer to Problem 4.64P

The equation of motion for the system is $\stackrel{⃛}{x}+\frac{k_1}{c}\ddot{x}+\frac{\left(k_1+k_2\right)}{m}\dot{x}+\frac{k_1{k}_2}{mc}x=\frac{k\dot{y}}{m}$.

Explanation of Solution

Figure (1) shows the free body diagram of the system.

Figure-(1)

Consider a point $A$ on the system and its displacement $x_A$ as shown in figure (1), the mass of the system is $m$, the damping coefficient is $c$, the stiffness of the spring (1) is $k_1$ and of the spring (2) is $k_2$, the displacement at different point are $x$, $y$, and $x_A$

Write the expression for horizontal force on mass $m$.

$m\ddot{x}=c\left({\dot{x}}_A-\dot{x}\right)-k_{2}x$ ..... (I)

Here, the acceleration is $\ddot{x}$, velocity at point $A$ is ${\dot{x}}_A$.

Write the expression for horizontal force at point $A$.

$k_1\left(y-x_A\right)=c\left({\dot{x}}_A-\dot{x}\right)$ ...... (II)

Write Laplace transform for Equation (I).

$\begin{array}{l}m{s}^{2}X\left(s\right)=\left[c\left(s{X}_A\left(s\right)-sX\left(s\right)\right)-k_{2}X\left(s\right)\right]\\ m{s}^{2}X\left(s\right)+k_{2}X\left(s\right)+csX\left(s\right)=cs{X}_A\left(s\right)\\ X\left(s\right)\left[m{s}^2+k_2+cs\right]=cs{X}_A\left(s\right)\\ \frac{X\left(s\right)\left[m{s}^2+k_2+cs\right]}{cs}=X_A\left(s\right)\end{array}$ ...... (III)

Write Laplace transform for Equation (II)

$\begin{array}{l}{k}_1\left[Y\left(s\right)-X_A\left(s\right)\right]=c\left[\left(s\left(X_A\left(s\right)-sX\left(s\right)\right)\right)\right]\\ k_{1}Y\left(s\right)-k_1{X}_A\left(s\right)=cs{X}_A\left(s\right)-csX\left(s\right)\\ k_{1}Y\left(s\right)+csX\left(s\right)=cs{X}_A\left(s\right)+k_1{X}_A\left(s\right)\end{array}$ ..... (IV)

Substitute $\frac{X\left(s\right)\left[m{s}^2+k_2+cs\right]}{cs}$ for $X_A\left(s\right)$ in Equation (IV)

$\begin{array}{l}{k}_{1}Y\left(s\right)+csX\left(s\right)=cs\frac{X\left(s\right)\left[m{s}^2+k_2+cs\right]}{cs}+k_1\frac{X\left(s\right)\left[m{s}^2+k_2+cs\right]}{cs}\\ k_{1}Y\left(s\right)+csX\left(s\right)=\left(cs+k_1\right)\left[\frac{m{s}^2+k_2+cs}{cs}\right]X\left(s\right)\\ k_{1}Y\left(s\right)=\left(cs+k_1\right)\left[\frac{m{s}^2+k_2+cs}{cs}\right]X\left(s\right)-csX\left(s\right)\\ k_{1}Y\left(s\right)=\left[\frac{\left(cs+k_1\right)\left(m{s}^2+k_2+cs\right)}{cs}-cs\right]X\left(s\right)\end{array}$

$\begin{array}{c}{k}_{1}Y\left(s\right)=\left[\frac{\left(cs+k_1\right)\left(m{s}^2+k_2+cs\right)}{cs}-cs\right]X\left(s\right)\\ k_{1}Y\left(s\right)=\left[\frac{mc{s}^3+c^2{s}^2+k_{2}cs+k_{1}m{s}^2+k_{1}cs+k_1{k}_2-c^2{s}^2}{cs}\right]X\left(s\right)\\ k_{1}Y\left(s\right)=\left[\frac{mc{s}^3+k_{2}cs+k_{1}m{s}^2+k_{1}cs+k_1{k}_2}{cs}\right]X\left(s\right)\end{array}$

$\begin{array}{l}{k}_{1}csY\left(s\right)=\left(mc{s}^3+k_{2}cs+k_{1}m{s}^2+k_{1}cs+k_1{k}_2\right)X\left(s\right)\\ s{k}_{1}cY\left(s\right)=mc{s}^{3}X\left(s\right)+k_{1}m{s}^{2}X\left(s\right)+s\left(k_{1}c+k_{2}c\right)X\left(s\right)+k_1{k}_{2}X\left(s\right)\end{array}$ ...... (V)

Take inverse Laplace of Equation (V).

$\begin{array}{l}{k}_{1}c\dot{y}=mc\stackrel{⃛}{x}+k_{1}m\ddot{x}+\left(k_{1}c+k_{2}c\right)\dot{x}+k_1{k}_{2}x\\ \stackrel{⃛}{x}+\frac{k_{1}m}{mc}\ddot{x}+\frac{\left(k_{1}c+k_{2}c\right)}{mc}\dot{x}+\frac{k_1{k}_2}{mc}x=\frac{k_{1}c\dot{y}}{mc}\\ \stackrel{⃛}{x}+\frac{k_1}{c}\ddot{x}+\frac{\left(k_1+k_2\right)}{m}\dot{x}+\frac{k_1{k}_2}{mc}x=\frac{k_1\dot{y}}{m}\end{array}$.

Conclusion:

The equation of motion for the system is $\stackrel{⃛}{x}+\frac{k_1}{c}\ddot{x}+\frac{\left(k_1+k_2\right)}{m}\dot{x}+\frac{k_1{k}_2}{mc}x=\frac{k\dot{y}}{m}$.