Mechanical Motion of Springs Differential Equations

  • #1

Homework Statement



Skjermbilde_2012_04_23_kl_6_50_05_PM.png


The Attempt at a Solution


So I've been interpreting the information in the problem as follows: [itex]F_{damping} = 4u' = μ(u')[/itex], [itex]k = \frac{4N}{m}[/itex]. If the system is critically damped then [itex]μ = 2\sqrt{km} = 2\sqrt{\frac{4N}{m}m} = 2\sqrt{4N}[/itex]. Now it seems as though the spring constant is cancelling out my mass, so if [itex]μ =4[/itex] then simply force [itex]N =1[/itex] 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

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
964
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.
 
  • #3
Let me see if I can do better with this now. I write,

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

Related Threads on Mechanical Motion of Springs Differential Equations

Replies
1
Views
3K
Replies
1
Views
1K
Replies
3
Views
3K
Replies
2
Views
2K
Replies
9
Views
2K
Replies
2
Views
9K
Replies
4
Views
680
  • Last Post
Replies
1
Views
1K
Replies
12
Views
2K
Top