# Mechanical Motion of Springs Differential Equations

## The Attempt at a Solution

So I've been interpreting the information in the problem as follows: $F_{damping} = 4u' = μ(u')$, $k = \frac{4N}{m}$. If the system is critically damped then $μ = 2\sqrt{km} = 2\sqrt{\frac{4N}{m}m} = 2\sqrt{4N}$. Now it seems as though the spring constant is cancelling out my mass, so if $μ =4$ then simply force $N =1$ and the system is critically damped for any mass. But then, this doesn't seem quite right. I must be misunderstanding the problem.

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
The "m" in N/m is NOT the mass. Nor is N a parameter. That is just saying that the spring constant is 4 Newtons per meter.

Let me see if I can do better with this now. I write,

Critically damped implies $μ = 2\sqrt{km}$. Given that $F_{damping} = μ(u') = 4u'$, then $μ = 4$ and we then say $4 = 2\sqrt{4m}$ which implies that m = 1.