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

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=
`(13* 32)/(y)=58.667lb/ft`

For critical damping
`\zeta =1`

`2* \zeta * \omega_n =(c)/(m)`

and
`\omega_n=\sqrt {(k)/(m)}`

`\omega _n=2.1241rad/s`

substituting values

`2* 1 * 2.124 =(c)/(m)`

c=55.227 lb.sec/ft