Answer:
Cc= 12.7 lb.sec/ft
Explanation:
Given that
m = 22 lb
g= 32 ft/s²
[tex]m = \dfrac{22}{32}=0.6875\ s^2/ft[/tex]
x= 4.5 in
1 in = 0.083 ft
x= 0.375 ft
Spring constant ,K
[tex]K=\dfrac{m}{x}=\dfrac{22}{0.375}[/tex]
K= 58.66 lb/ft
The damper coefficient for critically damped system
[tex]C_c=2\sqrt{mK}[/tex]
[tex]C_c=2\sqrt{0.6875\times 58.66}[/tex]
Cc= 12.7 lb.sec/ft
