Differential Equations A 2 kg mass on a spring with spring constant k = 1 is not oscillating. The system is critically damped. What is the coefficient of friction?

Question

asked Jan 12, 2020 89.2k views

1 Answer

Jizugu · Answer 1 · 2020-01-17T18:33:43+0000

Answer:

$b=\pm2√(2)$

Explanation:

For the spring mass system

my"+by'+ky=0

The system is said to be critically damped if
b^2-4mk=0

Here b is coefficient of friction

m is the mass of system

And k is spring constant

We have given m = 2 kg , k =1

So
b^2-4* 2* 1=0

b^2=8

$b=\pm2√(2)$

So the coefficient of friction is
$\pm2√(2)$