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## A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 new

Question

A spring with an mm-kg mass and a damping constant 9 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

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Physics
3 months
2021-08-09T11:58:19+00:00
2021-08-09T11:58:19+00:00 1 Answers
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## Answers ( )

Answer:4.05 kg

Explanation:From Hooke’s law, force required to stretch the spring is represented as follows:

k (0.5) = 2.5

Spring Constant, k = 5

For critical damping, c² – 4mk =0

m = c² / 4 k

c= damping constant

m = Mass to produce critical damping

There fore, m = (9²/4*5)

= 81/20

= 4.05 kg

Therefore, the mass that would produce critical damping. = 4.05 kg