Question

The relationship between the force f(t) and the displacement x(t) of a spring-mass system with a mass M, viscous damping D and spring constant K,

$M\frac{d^{2}x\left(t\right)}{d{t}^2}+D\frac{dx\left(t\right)}{dt}+Kx\left(t\right)=f\left(t\right)$

X(s) and F(s) are the Laplace transforms of x(t) and f(t) respectively. With M= 0.1, D = 2, K= 10 in appropriate units, the transfer function
$G\left(s\right)=\frac{X\left(s\right)}{F\left(s\right)}\text{is}$

• A. $\frac{10}{s^2+20s+100}$
• B. $s^2+20s+100$
• C. $\frac{10s^2}{s^2+20s+100}$
• D. $\frac{s}{s^2+20s+100}$

Answer 1

The correct option is A $\frac{10}{s^2+20s+100}$

$M\frac{d^{2}x\left(t\right)}{d{t}^2}+D\frac{dx\left(t\right)}{dt}+Kx\left(t\right)=f\left(t\right)$

Taking Laplace transform on both sides,

$M{s}^{2}X\left(s\right)+DsX\left(s\right)+KX\left(s\right)=F\left(s\right)$

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{1}{M{s}^2+DS+K}$

By substituting, M, D and K values,

$\frac{X\left(s\right)}{F\left(s\right)}=\frac{1}{0.1s^2+2s+10}$

$=\frac{10}{s^2+20s+100}$