2nd order mass, spring damper in series

In summary, the conversation discusses deriving differential equations for a mass attached to a spring and damper system, where the natural length of the spring is not taken into account. The equation for the system is derived and questions about the reference points and definitions of variables are addressed. Part 2, which involves a more complicated system, is also briefly discussed.
  • #1
hihiip201
170
0
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.
 
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  • #2
hihiip201 said:
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 = kxs(t).

You're just arbitrarily calling x = 0 the point of the mass at which the spring is relaxed and xs = 0. Note that xs has its own reference point, i.e. at the left-hand end of the spring. So the length of the spring is immaterial. If xs = 0 always there would be no excitation and no motion.
 
  • #3
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##.
 
  • #4
rude man said:
Your equation is correct. Call v = dx/dt and rewrite as m d2x/dt2 + k(x) + Bdx/dt = kxs(t).

You're just arbitrarily calling x = 0 the point of the mass at which the spring is relaxed and xs = 0. Note that xs has its own reference point, i.e. at the left-hand end of the spring. So the length of the spring is immaterial. If xs = 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!
 
  • #5
jhae2.718 said:
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!
 
  • #6
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".
 
  • #7
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"??
 
  • #8
rude man said:
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.
 

Related to 2nd order mass, spring damper in series

1. What is a 2nd order mass, spring damper in series system?

A 2nd order mass, spring damper in series system is a mechanical system consisting of a mass, a spring, and a damper connected in series. It is used to model a wide range of physical systems such as car suspensions, vibrating systems, and shock absorbers.

2. What is the equation for the motion of a 2nd order mass, spring damper in series system?

The equation for the motion of a 2nd order mass, spring damper in series system is given by the second-order differential equation:
m * d^2x/dt^2 + b * dx/dt + k * x = F(t)
where m is the mass, b is the damping coefficient, k is the spring constant, x is the displacement, and F(t) is the external force applied.

3. How does the stiffness of the spring affect the behavior of the system?

The stiffness of the spring determines how much force is required to stretch or compress the spring. A higher stiffness will result in a higher natural frequency and faster oscillations of the system. Conversely, a lower stiffness will result in a lower natural frequency and slower oscillations of the system.

4. What is the role of the damping coefficient in the system?

The damping coefficient represents the resistance of the system to motion. It dissipates energy from the system and helps to reduce the amplitude of oscillations, resulting in a smoother and more controlled motion. A higher damping coefficient will lead to a faster decay of oscillations, while a lower damping coefficient will result in longer-lasting oscillations.

5. How can the response of a 2nd order mass, spring damper in series system be analyzed?

The response of the system can be analyzed by solving the differential equation using mathematical methods such as Laplace transforms or numerical methods. The system's natural frequency, damping ratio, and transient and steady-state responses can be determined from the solution. Additionally, various techniques such as frequency response analysis and modal analysis can also be used to analyze the system's behavior.

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