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3 years ago

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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.

1 answer:

Dominik [7]3 years ago

5 0

Answer:

Cc= 12.7 lb.sec/ft

Explanation:

Given that

m = 22 lb

g= 32 ft/s²

$m = \dfrac{22}{32}=0.6875\ s^2/ft$

x= 4.5 in

1 in = 0.083 ft

x= 0.375 ft

Spring constant ,K

$K=\dfrac{m}{x}=\dfrac{22}{0.375}$

K= 58.66 lb/ft

The damper coefficient for critically damped system

$C_c=2\sqrt{mK}$

$C_c=2\sqrt{0.6875\times 58.66}$

Cc= 12.7 lb.sec/ft

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