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Mass Spring Damper System Model

  1. Jan 11, 2009 #1
    I am trying to write a system model for lowering a mass into water. In my system model, I have a mass hanging freely from a spring and damper connected in parallel. I then try to equate the damping coefficient to the drag experienced by the mass.

    I have provided the system equation below:

    my'' = mg - by' - ky where by' = 0.5*rho*Cd*A*y'^2 (drag)

    the full equation is then:

    my'' = mg - 0.5*rho*Cd*A*y'^2 - ky

    My questions:

    1) Having the spring and damper in parallel creates a second order non-linear differential equation (atleast I think)...which I have no idea howto solve. I have solved both 2nd order linear and 1st order non-linear diff eq but not 2nd order non-linear. Infact my notes said solving them is kind of a crap shoot. I have found examples with the spring and damper in series rather than parallel, but this does not make sense to me as the spring and damping act together to slow the mass. Is my model correct using parallel instead of series between the spring and damper?

    2) If so, is it ok to equate the damping force to the drag force inorder to find a damping coefficient?

    Any and all help is greatly appreciated.
     
    Last edited: Jan 12, 2009
  2. jcsd
  3. Jan 13, 2009 #2
    You really cannot say
    by' = 0.5*rho*Cd*A*y'^2
    but there is no need to do so.
    Your last equation is not too bad a model, although I might ask a couple of question about it.
    1) Where does the spring rate come from? Is that rope stretch?
    2) You do not appear to have taken buoyancy into account. As the body enters the water, this will add another term to the equation.
     
  4. Jan 13, 2009 #3
    Dr. D,

    Thanks for your reply! I was starting to worry noone would answer.

    I am a little confused by your response - You say that I can't really equate by' = 0.5*rho*Cd*A*y'^2. The confusion comes in your next, seemingly contradictory statement, "but there is no need to do so...the last equation is not too bad of a model." Did you mean I CAN equate the two, but there is no reason to do so?


    To answer your questions:

    My spring does indeed come from the stretch of a steel cable (k = EA/L). The reason I initially equated by' = 0.5*rho*Cd*A*y'^2was to determine a damping coefficient (since I don't know a typical value, b, for damping coefficient in seawater). It seems to me that the drag and damping forces are equal, as all damping effects are caused by the water. Bouyancy is something I need to add and was actually an oversight on my part, thank you for mentioning that. Would the bouyant force just be tacked on as another constant? I know bouyant force, Fb = rho*g*displaced Vol, which has no dependency on y.

    I really hope your response implies the damping coefficient is much simpler than what I am proposing. The by' = 0.5*rho*Cd*A*y'^2 equation creates a non-linearity in my system equation.
     
    Last edited: Jan 13, 2009
  5. Jan 13, 2009 #4
    Do you have a Free Body Diagram for this system? If not, you should really stop and make one right now. It should show the mass and all the forces acting on it. The forces involved will be
    The upward force of the spring (rope)
    The downward force of the weight
    The upward force of buoyancy (due to weight of water displaced) with will be a function of the depth of immersion
    The damping force

    Now sum the forces, taking positive upward:
    Sum F = Fspring + Fbuoy - W - Fdamp
    where I have assumed that the mass is partially submerged and moving upward (which gives Fdamp directed down).

    Now your biggest dilemma seems to be what to do about the damping force. If the motion is relatively slow, then a b*ydot term is probably a good description and this will give you a linear differential equation. The square law drag applies more to high speed drag.

    You also need the buoyant term and that will be equal to the weight of the water displaced. I don't quite know how you have your coordinates set up, so I don't know how to tell you to express this term. It will be a good thing for you to work through this, however. If the mass is of constant section area in the water plane, this will give you a second term linear in the displacement, just like another spring term.

    Give all of that a try.
     
  6. Jan 13, 2009 #5
    I do have a freebody diagram as described, however, I am lowering the mass therefore my damping term will be opposite yours. FYI, my coordinate system is setup positive down
    + weight
    + inertia
    - bouyant force
    - damping force
    - spring force

    m y'' = W - Fdamp - Fspring - Fbouyant

    the motion is relatively slow which is why I have the by' term, however I am lost with where I get damping coefficient b from. I have search online without any luck of finding any tabulated or experimental values of damping coefficient in seawater.

    I agree that the bouyant term is a function of depth immersed; however, I think it is safe to assume the body is fully immersed, creating a constant bouyant force term (the weight of the water displaced is the weight of the water of the entire masses volume, always). correct?


    Thanks again for your help, it is really appreciated!
     
    Last edited: Jan 13, 2009
  7. Jan 13, 2009 #6
    If the body is fully immersed, then there is the question, does it float? Is there still tension in the supporting rope? If it is fully immersed, yes, the buoyant force is a constant.

    As to finding a value for the coefficient b, that will be tough. It will depend on the shape of the body, on the wetted surface area, and the phase of the moon. In other words, good luck with that. You might want to think about an experiment. It probably will not be large. There will probably be more damping due to other sources than the water friction, such as internal damping in the rope, in pulleys, etc. Friction terms are very hard to evaluate in real life.
     
  8. Jan 13, 2009 #7
    Dr. D,

    I am assuming a heavy object which will not float and yes tension in the rope.

    I think my questions have been answered. I was really concerned with obtaining a value for damping coefficient, b...as you stated would probably need an experiment. Would you happen to know general or typical magnitude of common damping coefficients (are we talking .01, 1, or 10, etc)?

    Just for learning sake, if the mass were moving at high speed...would the drag square law substitution I made in the system equation be appropriate? Then I would be left with a 2nd order non-linear diff eq.
     
  9. Jan 14, 2009 #8
    As I said earlier, "The square law drag applies more to high speed drag."

    I have no way to give you a value for b. I don't even know what system of units you are using, but even if I did, it would be only a wild guess. Please make your own guess.
     
  10. Jan 14, 2009 #9
    Thank you for your help Dr. D.
     
  11. Jan 16, 2009 #10
    When examining the damping in some fluid, we have to take account of the flow around the oscillating object. We will see that this can be taken into account as a "change" of mass of the object, thus we have to consider an effective mass.


    So first of all lets find the flow (velocity field) around the object. Let the object be a sphere (So we dont get "messy"... :D).

    For simplicity, lets assume that the fluid is ideal (this is a reasonable approximation, given the circumstances), that is frictious forces can be neglected, it is also incompressible (another reasonable assumption), and the flow can be considered as potential flow.

    So the above assumptions can be collected into to equations.

    Since the fluid is incompressible (the mass density is constant throughout the fluid), due to the continuity equation we have:

    [tex]\text{div}\textbf{v}=0[/tex]

    We also assumed potential flow, which means that the curl of the velocity field vanishes. This means that the velocity field can be given as the gradient of a scalar function [tex]\Phi[/tex], called the velocity potential [tex]\textbf{v}=\nabla \Phi[/tex]:

    [tex]\text{curl}\textbf{v}=0 \Longrightarrow \textbf{v}=\text{grad}\Phi[/tex]

    Combining these two equations we have:

    [tex]\text{div}\textbf{v}=\text{div}\text{grad}\Phi=\bigtriangleup \Phi=0[/tex]

    That is we obtained a Laplace equation for the velocity potential. We also need the boundary conditions. These depend on the geometry of the object in the flow.

    Now consider a sphere of radius R travelling with uniform velocity [tex]\textbf{u}[/tex] in the fluid. Now we want the velocity of the fluid to vanish at infinity. So we need a solution of the laplace equation which vanishes at infinity. We know this to of the form [tex]1/r[/tex], or higher order derivatives of this. Due to the symmetry of the system, the only vector the velocity potential can depend on one constant vector, the velocity [tex]\textbf{u}[/tex] of the sphere. Now we can construct a scalar by taking the inner product of some vector [tex]\textbf{A}[/tex] and the gradient of 1/r.
    Now we can make further assumptions on the u velocity dependence of the vector A.

    Lets consider the boundary conditions. The boundary conditions, for the potential flow of an ideal fluid around an object, are that the normal component of the flow at a given point be equal to the normal component of the velocity of the object at that point.

    That is at [tex]r=R[/tex] we have:

    [tex]\textbf{u}\textbf{n}=\textbf{v}\textbf{n}[/tex]

    Where n is a unit vector normal to the surface of the sphere.

    So we see that the boundary conditions are linear in [tex]u[/tex]. Combining this with our previous analysis, means that velocity potential can contain u only linearly. So the velocity potential will be:

    [tex]\Phi=\textbf{A}\cdot\text{grad}\left(\frac{1}{r}\right)[/tex]

    Taking the gradient of this we get the velocity field:

    [tex]\textbf{v}=\text{grad}\Phi = \frac{3(\textbf{A}\textbf{n})}{r^3}-\frac{\textbf{A}}{r^3}[/tex]

    Now we only have to find the value of A. Using the boundary conditions we have:

    [tex]\left(\frac{3(\textbf{A}\textbf{n})}{r^3}-\frac{\textbf{A}}{r^3}\right)\textbf{n}\bigg |_{r=R}=2\frac{\textbf{A}\textbf{n}}{R^3}=\textbf{u}\textbf{n}[/tex]

    So:

    [tex]\textbf{A}=\frac{R^3}{2}\textbf{u}[/tex]

    Plugging this back into the previous equation, the velocity field around the sphere will be:

    [tex]\textbf{v}=\frac{R^3}{2r^3}\left(3(\textbf{u}\textbf{n})\textbf{n}-\textbf{u}\right)[/tex]

    Now to find the effect of the flow on the movement of the sphere, lets consider its kinetic energy. We have to calculate the kinetic energy of the fluid flowing around the sphere, and this can be interpreted as if the object gained some "extra" mass, hence the frequency of its oscillations have shifted.

    The kinetic energy of the fluid:

    [tex]E_k=\int \frac{\rho}{2}v^2 \; dV[/tex]

    Where the volume integral is taken for the "whole" fluid. and \rho is the mass density of the fluid.

    Taking into account the symmetry of the system, we introduce spherical coordinates, where the z axis, corresponds to the direction of the velocity u, [tex]\theta[/tex] the polar angle (that is the angle between u and the normal unit vector n). So we have [tex]\textbf{u}\textbf{n}= u\cos\theta[/tex]

    (continued in the next post)
     
  12. Jan 16, 2009 #11
    With this the needed square of the velocity field is:

    [tex]v^2=\frac{R^6}{4r^6}\left(9(\textbf{u}\textbf{n})^2-6(\textbf{u}\textbf{n})^2+u^2\right)=\frac{R^6}{4r^6}\left(3u^2\cos^2\theta+u^2\right)[/tex]

    The volume element in spherical coordinates is: [tex]dV= r^2\sin\theta\;d\theta;d\varphi\;dr[/tex]

    So finally the kinetic energy integral is:

    [tex]E_k=\frac{\rho}{2}\int_R^{\infty}dr\;r^2\int_0^{\pi}d\theta\;\sin\theta\int_0^{2\pi}d\varphi\;\frac{R^6}{4r^6}\left(3u^2\cos^2\theta+u^2\right)[/tex]

    Performing the integration, we get:

    [tex]E_k=\frac{R^3\pi}{3}\rho u^2= \frac{1}{2}\underbrace{\left(\frac{2R^3\pi}{3}\rho\right)}_{m'}u^2[/tex]

    So as we see this can truly be interpreted as if the original mass [tex]m[/tex] of the sphere grew by [tex]m'=\frac{2R^3\pi}{3}\rho[/tex].

    So the total "effective" mass M of the spherical object will be:

    [tex]M=m+m'=m+\frac{2R^3\pi}{3}\rho[/tex]

    Now we can just plug this effective mass into the usual damped oscillation equation, and we get the answer.

    Finally lets summarize, the conditions in which our above model is appropriate.
    First of all we neglected all frictious, viscosous effects in the fluid i.e. we considered the fluid to be ideal. This assumption can be made if you are working in water or other materials with low viscosity, but obviously cannot be made when you are in for eg. honey. In the viscous case we have to work harder, and can obtain some sort of velocity profile under the assumptions of stokes flow, that is we have to solve the linearized navier stokes equations. In the case of the sphere this is can be done fairly easily in a similar manner as above in finite time.
    We also assumed potential flow. i.e. the velocity field is irrotational.

    And most importantly all of these assumptions can be satisfied if the velocity of the oscillating object is not so great.
     
  13. Jan 16, 2009 #12
    Thaakisfox,

    Wow. I totally wasn't anticipating a response like this. This may take me a while to digest (goes back to a viscous fluid flow class I took a while ago)...I see in the end you derived an effective mass for the spring system. Am I correct in saying this effective mass would take place of the dampening term in the system equation and I would be left with only the spring term?

    as in:

    M y'' = Mg - ky where M is the effective mass
     
  14. Jan 16, 2009 #13
    The original problem statement said nothing about an "oscillating sphere" but it did say "for lowering a mass into water." The added mass rountine is significant if we truly want to talk about the rapid oscillations of a body in water; it is not significant, and not worth all the extra effort, if we are talking about the very slow motions of a body being slowly lowered into the water on an elastic structure.

    This is an example of "problem creep" that seems to occur often on PF. When the original problem statement is expanded by a later answer that makes a much bigger and grander problem out of the situation than was ever indicated in the original problem. Is this useful? I do not see how it is. It certainly gives an opportunity to show off, but does it really help, or does it just confuse?

    For someone who is having difficulty writing the initial equations for this system, it is unlikely that they are really looking for an added mass solution. Perhaps that is exactly what was sought, but it certainly was not indicated in the original request. The request was for a damping coefficient, with emphasis on a velocity related term. The added mass response really seems like overkill.
     
  15. Jan 16, 2009 #14
    Dr. D,

    Thakkisfox responded to another thread I had created (before any responses were posted on this one). I then referred him to this discussion.


    His response was:
    "Your equation is correct :D
    It is basically a 1D motion, of a damped driven harmonic oscillator..

    But if you want to do it in sea water, then it wont be correct, since you have to take into account the flow around the hanging mass, which is not so easy. For a sphere it is easy to calculate the frequency shift due to the flow, because of the symmetry, but for another assymetric object its not so easy..."

    I sent him a message thanking him for his help and stated that I was lowering simply shaped objects (sphere, flat plate, prism). Although I wasn't expecting a derivation, this was his response to that.

    I don't want this to turn into an argument, but I don't think he was trying to show off...rather help me.

    Thank you both for your time and responses. They were tremendously helpful.
     
  16. Jan 16, 2009 #15
    Actually I dont see, what is the overkill in the answer. The poster originaly posted this in another section of PF, and wanted to do the experiment in sea water. Hence I replied in the original post, that we could consider the flow. The poster of the problem then wrote me a PM, that he was experimenting with simple geometry, sphere, cube etc. and asked me to provide some indepth analysis here in the engineering forum..., so this was my analysis I did, its not "an anshow off", Its just alysis of the problem with some assumptions.., then also we can consider different kinds of rapid oscillations.., that would be a more difficult analysis...

    You yourself said that

    "For someone who is having difficulty writing the initial equations for this system, it is unlikely that they are really looking for an added mass solution. Perhaps that is exactly what was sought, but it certainly was not indicated in the original request."

    If the poster is not familiar with the outline of the problem, then it is our duty to make it clear. This is not a Homework problem, so I guess giving a bit more than just blabbered sentences is expected.
    So how could the original request contain something that is not known by the author, but is an important part of the description...
     
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