Question

A mass weighing 13 lb stretches a spring 4.5 in. The mass is also attached to a damper with coefficient Y. Determine the value of Y for which the system is critically damped. Assume that g = 32 ft/s^2.

Answer 1

Answer:55.227

Step-by-step explanation:

Given data

mass $\left ( m\right )$=13lb

spring stretches by $\left ( x\right )$=4.5in=0.375 ft

g=$32ft/s^2$

Now Spring constant K

mg=kx

k=$\frac{13\times 32}{y}=58.667lb/ft$

For critical damping $\zeta =1$

$2\times \zeta \times \omega_n =\frac{c}{m}$

and $\omega_n=\sqrt {\frac{k}{m}}$

$\omega _n=2.1241rad/s$

substituting values

$2\times 1 \times 2.124 =\frac{c}{m}$

c=55.227 lb.sec/ft