Question

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?

Answer 1

`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)`