2nd order mass, spring damper in series

[hihiip201]

1. Homework Statement

B, K, M

2. Homework Equations

1. xs(t) -----spring ----mass-----damper-----fixed, derive DE for x of mass


given :2. F - > M -----spring-------damper ---- fixed in series, derive the DE for velocity of spring


3. The Attempt at a Solution


1. ma = -k(x-xs) - B(v)

But I don't understand, why aren't we taking the natural length of the spring into account?



2. no idea, I have the solution and it says that damping force is B(v(t) - vk) , but I have no idea what v has anything to do with damper, the relative velocity vb should just be vk - 0 to me but the textbook and the homework solution suggest otherwise.

 

[rude man]

1. 3. The Attempt at a Solution


1. ma = -k(x-xs) - B(v)

But I don't understand, why aren't we taking the natural length of the spring into account?

.

Your equation is correct. Call v = dx/dt and rewrite as m d²x/dt² + k(x) + Bdx/dt = kx_s(t).

You're just arbitrarily calling x = 0 the point of the mass at which the spring is relaxed and x_s = 0. Note that x_s has its own reference point, i.e. at the left-hand end of the spring. So the length of the spring is immaterial. If x_s = 0 always there would be no excitation and no motion.

 

[jhae2.718]

Note that a lot of analysis of dynamical systems is of perturbations from the equilibrium position, so the natural length of the spring will fall out in those cases.

There are also "mathematical springs" where the response is defined as $kx$.

 

[hihiip201]

Your equation is correct. Call v = dx/dt and rewrite as m d²x/dt² + k(x) + Bdx/dt = kx_s(t).

You're just arbitrarily calling x = 0 the point of the mass at which the spring is relaxed and x_s = 0. Note that x_s has its own reference point, i.e. at the left-hand end of the spring. So the length of the spring is immaterial. If x_s = 0 always there would be no excitation and no motion.

ya after thinking about that for like 5 minutes on the exam I finally figured out, thank you very much!


And I think I know what happens with vk, so vk is probably defined to be vm - v2 where 2 is the point to the right of spring and left of damper, so they are saying Bv2, but expressing v2 in terms of vm - vk.


I hate it when TA just throw variables at me without explaining, or maybe I should have remembered the definition of vk sooner.


thank you very much!

 

[hihiip201]

Note that a lot of analysis of dynamical systems is of perturbations from the equilibrium position, so the natural length of the spring will fall out in those cases.

There are also "mathematical springs" where the response is defined as $kx$.

understood, thank you!

 

[rude man]

Part 2 is more difficult. That's because the spring's left and right boundaries are not defined. Also, I'm not sure what is meant by "the velocity of the spring".

 

[rude man]

I mean, the left & right ends of the spring are going to move at different velocities, so what is meant by 'the velocity of the spring"??

 

[hihiip201]

I mean, the left & right ends of the spring are going to move at different velocities, so what is meant by 'the velocity of the spring"??

sorry i think vk is defined to be the difference of each end of the spring, not velocity of the spring.