Chapter 4, Problem 4.90P

To determine

(a)

The equations of motion of the system shown in Figure P4.90.

Answer to Problem 4.90P

The equations of motion of the system given below,

$f+9\times {10}^4{x}_1-6\times {10}^4{x}_2=20\stackrel{••}{x_1}$

$6\times {10}^4(x_2-x_1)=60\stackrel{••}{x_2}$.

Explanation of Solution

Given:

The masses are $m_1=20\text{ kg}$ and $m_2=60\text{ kg}$

The spring constant are $k_1=3\times {10}^4\text{ N/m and }{k}_2=6\times {10}^4\text{ N/m}$

Concept used:

Newton’s Second law of motion and Hooke’s Law are used to obtain the equation of motions.

Figure P4.90

Calculation:

Free body diagram of $m_1$ and $m_2$ are given below,

Applying Newton’s Second law of motion to mass m₁

$f-k_1{x}_1-k_2(x_1-x_2)=m_1\stackrel{••}{x_1}$

$k_2(x_1-x_2)=m_2\stackrel{••}{x_2}$

Substituting values for $k_1,k_2{\text{ and m}}_1$

$f-3\times {10}^4{x}_1-6\times {10}^4(x_1-x_2)=20\stackrel{••}{x_1}$

$f-9\times {10}^4{x}_1+6\times {10}^4{x}_2=20\stackrel{••}{x_1}$

Substituting values for $k_2{\text{ and m}}_2$

$6\times {10}^4(x_1-x_2)=60\stackrel{••}{x_2}$

Therefore the equations of motion of the system given below,

$f-9\times {10}^4{x}_1+6\times {10}^4{x}_2=20\stackrel{••}{x_1}$ (1)

$6\times {10}^4(x_1-x_2)=60\stackrel{••}{x_2}$ (2)

Where,

$f$ = Applied force to mass m₁$k_1,k_2$ = Stiffness of the spring

$x_1,x_2$ = Displacement of the spring respectively

$m_1,m_2$ = Mass of the objects.

To determine

(b)

The transfer functions for $X_1(s)/F(s),X_2\left(s\right)/F(s)$.

Answer to Problem 4.90P

$\frac{X_1\left(s\right)}{F\left(s\right)}=\frac{s^2+1000}{20s^4+11\times {10}^4{s}^2+3\times {10}^7}$

$\frac{X_2\left(s\right)}{F\left(s\right)}=\frac{1000}{\left(20s^4+11\times {10}^4{s}^2+3\times {10}^7\right)}$.

Explanation of Solution

Given:

From part (a),

$f-9\times {10}^4{x}_1+6\times {10}^4{x}_2=20\stackrel{••}{x_1}\to \left(1\right)$

$6\times {10}^4(x_1-x_2)=60\stackrel{••}{x_2}\to \left(2\right)$

Concept used:

Laplace Transformation used to obtain the transfer functions.

Assume zero initial conditions for $x_1\text{ and }{x}_2$.

Calculation:

Converting the equations (1) and (2) to Laplace domain,

$F(s)-9\times {10}^4{X}_1(s)+6\times {10}^4{X}_2\left(s\right)=20\left(s^2{X}_1\left(s\right)-\stackrel{•}{x_1}\left(0\right)-s{x}_1\left(0\right)\right)\to \left(3\right)$

$6\times {10}^4\left(X_1\left(s\right)-X_2\left(s\right)\right)=60\left(s^2{X}_2\left(s\right)-\stackrel{•}{x_2}\left(0\right)-s{x}_2\left(0\right)\right)\to \left(4\right)$

for zero initial condition, $x_1\left(0\right)=0\text{ , }{x}_2\left(0\right)=0\text{ , }\stackrel{•}{x_1}\left(0\right)=0\text{ , }\stackrel{•}{x_2}\left(0\right)=0$

$F(s)-9\times {10}^4{X}_1(s)+6\times {10}^4{X}_2\left(s\right)=20s^2{\text{X}}_1\left(s\right)\to \left(5\right)$

$6\times {10}^4\left(X_1\left(s\right)-X_2\left(s\right)\right)=60s^2{X}_2\left(s\right)\to \left(6\right)$

from equation (6),

$X_2\left(s\right)=\frac{1000}{s^2+1000}{X}_1\left(s\right)\to \left(7\right)$

From equation (7) and (5),

$\frac{X_1\left(s\right)}{F\left(s\right)}=\frac{s^2+1000}{20s^4+11\times {10}^4{s}^2+3\times {10}^7}$$\to \left(6\right)$

From equation (7) and (6),

$\frac{X_2\left(s\right)}{F\left(s\right)}=\frac{1000}{\left(20s^4+11\times {10}^4{s}^2+3\times {10}^7\right)}$$\to \left(7\right)$.

To determine

(c)

A plot of unit step responses of $x_1$ for zero initial conditions.

Answer to Problem 4.90P

Explanation of Solution

Given:

unit-step response of $x_1$ for zero initial conditions

$\frac{X_1\left(s\right)}{F\left(s\right)}=\frac{s^2+1000}{20s^4+11\times {10}^4{s}^2+3\times {10}^7}$

Solution:

Using MATLAB,

>>sys1=tf([1,0,1000],[20,0,11*10^4,0,3*10^7]);

>> step(sys1)