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alexira [117]
3 years ago
11

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 o

f Y for which the system is critically damped. Assume that g = 32 ft/s^2.
Mathematics
1 answer:
Gemiola [76]3 years ago
5 0

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

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