Question

A mass weighing 22 lb stretches a spring 4.5 in. The mass is also attached to a damper with Y coefficient . Determine the value of Y for which the system is critically damped. Assume that g=32 ft/s2. Round your answer to three decimal places.

Answer 1

Cc= 12.7 lb.sec/ft

Step-by-step explanation:

Given that

m = 22 lb

g= 32 ft/s²

`m = (22)/(32)=0.6875\ s^2/ft`

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

`K=(m)/(x)=(22)/(0.375)`

K= 58.66 lb/ft

The damper coefficient for critically damped system

`C_c=2√(mK)`

`C_c=2√(0.6875* 58.66)`

Cc= 12.7 lb.sec/ft