Answer:55.227
Step-by-step explanation:
Given data
mass [tex]\left ( m\right )[/tex]=13lb
spring stretches by [tex]\left ( x\right )[/tex]=4.5in=0.375 ft
g=[tex]32ft/s^2[/tex]
Now Spring constant K
mg=kx
k=[tex]\frac{13\times 32}{y}=58.667lb/ft[/tex]
For critical damping [tex]\zeta =1[/tex]
[tex]2\times \zeta \times \omega_n =\frac{c}{m}[/tex]
and [tex]\omega_n=\sqrt {\frac{k}{m}}[/tex]
[tex]\omega _n=2.1241rad/s[/tex]
substituting values
[tex]2\times 1 \times 2.124 =\frac{c}{m}[/tex]
c=55.227 lb.sec/ft
