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Homework Help: Diff-EQ Problem - Two Springs

  1. Mar 27, 2009 #1
    From "Differential Equations" 3rd Edition, Blanchard/Devaney/Hall
    Section 2.3, Question #13

    I have the final answer from the text, as well as my Professors take (class has 7 people).
    Looking for a more thorough explanation. Work has already been graded, but I still don't get it completely.

    From the text:
    We consider a mass sliding on a frictionless table between two walls that are 1-unit apart and connected to both walls with springs, as shown below.

    |-////-M-\\\\-| (M is the mass... hope you get the idea from this crude "drawing")

    Let k1 and k2 be the spring constants of the left and right spring, respectively, let m be the mass, and let b be the damping coefficient of the medium the spring is sliding through. Suppose L1 and L2 are the rest lengths of the left and right springs, respectively.

    Write a second-order differential equation for the position of the mass at time t. [Hint: The first step is to pick an origin, that is, a point where the position is 0. The left hand wall is a natural choice.] (Incidentally, our professor wants us to choose the left hand wall as the origin).

    My attempt, my professors explanation, and the answer.
    My first thought was that we don't know if the springs in the drawing are compressed, under tension, or whether their lengths at rest are in excess of 1-unit. My professor said that the text is written by mathematics professors, and we are to assume that L1+L2 are less than 1-unit in length.

    With that I proceeded to set up a force equation.

    m(dx2/dt2)+b(dx/dt)+k1x+k2x = "Something"

    Turns out the "Something" is k1L1-k2(L2+1)

    I understand that this is the force experienced by the stationary block when pressed against the left wall (when x=0), but what is this equation really stating?

    Reading it all, I'm interpreting it as "The force due to the accelerated mass plus the drag/frictional force (actually minus) plus the spring force of spring one plus (actually minus) the spring force of the other spring is equal to the sum of the spring forces". For some reason I'm not getting it conceptually.

    Please help if you can.
    Last edited: Mar 27, 2009
  2. jcsd
  3. Mar 27, 2009 #2


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    Try writing down all the forces at play here, without the differentials to get a clearer picture of what is going on. As one spring is compressed, what does the other one do? (this will give you a hint as to the sign of k in each term). Try drawing a free body diagram.

    Edit: also its worth noting that the "Something" is the forcing function of the system. Why would the springs have a role in the forcing function? Think of the forces at play when the mass oscillates.
    Last edited: Mar 27, 2009
  4. Mar 27, 2009 #3
    Well we have four forces here (I believe). The force due to the acceleration of the mass, the force due to drag, and the two spring forces. The force due to acceleration of the mass is responsive, and acts in the opposite direction of the initial applied force. The force due to drag is opposite that of the mass' motion, and the spring force directions depend on the mass' position.

    When held against the left wall and then released, the force due to acceleration of the mass and drag force act to the left, and both spring forces act to the right (the left pushing and the right pulling).

    I'm still not getting the Diff-EQ in it's entirety... it's really irritating :frown:
  5. Mar 27, 2009 #4


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    There is no "force due to the acceleration. There are three forces- the forces due to each of the springs and the damping force. When the mass is pressed to the left side and released, yes both spring forces are to the right and the damping force is to the left but that will not be true in general. The direction of the spring force depends upon the postion of the mass and the direction of the damping force depends on direction of motion at each moment.
  6. Mar 27, 2009 #5
    There is no force due to the acceleration of the mass? I thought thats what the m(d2x/dt2) term was? F=ma.

    It's a term in the damped harmonic oscillator equation. My text says that this term is indeed a force, in addition to the dampening (friction/drag) force b(dx/dt) and any forces from springs (using Hooke's Law).

    So now I'm really confused if the text is wrong. :frown:
  7. Mar 27, 2009 #6


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    That's the resultant force.
  8. Mar 27, 2009 #7
    So could you explain/walk-through what the equation states in words? I'm having a really difficult time understanding the answer.

    The text & my professor give the following as the answer...
    m(d2x/dt2)+b(dx/dt)+x(k1+k2) = k1L1-k2(L2+1)
  9. Mar 27, 2009 #8


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    You would have to rewrite it to state that, the resulting force ma is equal to the force of friction/damping ([tex]-b \frac{dx}{dy}[/tex]) plus the restoring forces of spring 1 and 2 ([tex]-k_1x-k_1x[/tex]). However in motion, the spring-mass system is forced by the springs themselves and obeys [tex]k_1L_1-k_2(L_2+1)[/tex]
  10. Mar 27, 2009 #9
    Thanks for the response.

    I get the equation m(d2x/dt2) = -b(dx/dt)-k1x-k2x.

    I guess it's the other part that is confusing me. I don't understand your explanation for that part.
  11. Mar 27, 2009 #10


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    Try a thought experiment of every instant the mass is oscillating. Each time one spring is compressed and the other is stretched how are the forces affecting the oscillation? the left side of the differential equation you just wrote only models the oscillation alone. Now how is the oscillation affected?
  12. Mar 27, 2009 #11
    I'm feeling dumb here... somewhere I'm not understanding something.
    Shamefully I have about 8 hours of thinking into this today.

    The way I'm thinking of this (and obviously I'm wrong somewhere) is a system with two spring forces that act opposite one another on a mass. When one enters compression it exerts a force in the opposite direction, while the one under tension pulls towards itself. The movement of the mass faces a friction/drag resistance through the medium it travels through which is against the direction of movement and is proportional to velocity. Another force in the direction of movement of the accelerated mass increases as the mass is accelerated and decreases as the mass is decelerated (by the drag, and push/pull relationship of the springs). When the mass is decelerated to a zero velocity, it begins traveling in the opposite direction due to the spring forces - oscillating.

    Where am I going awry... :confused:
  13. Mar 28, 2009 #12
    my attempt, after reviewing wikipedia "harmonic oscillator"...

    mx'' is newtons second law: the sum of all the forces is proportional to mx'', so we want to sum up all the forces and set them equal to this.

    mx''=kx is a simple harmonic oscillator, where kx is the restoring force in the spring. there are two springs so i would want two terms. [edit] it looks like these are additive rather than cancelling. linearly proportional to position, so partial compressions dont affect the outcome. (i hope?)

    if you add in a cx' term, this is a dampening term, dissipation into the walls rather than friction, i think. [edit] friction is generally vague to model, so this is all we can get.

    finally you can add in a driver function based on the model. i'm not sure where to get this though.

    close enough?
    Last edited: Mar 28, 2009
  14. Mar 28, 2009 #13
    I think it's the "driver function" part I'm not getting.
  15. Mar 28, 2009 #14


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    Nick, try looking up forcing functions. Usually forcing/driving functions are external to the system, for instance if you were to push a rocking chair once and let it go oscillate that would be the left side of the diff eq. Now suppose instead you decided to keep rocking the chair, i.e. forcing its oscillations in some pattern you desired-that would be modeled as the right side of the diff eq. You could force it in a timely manner (periodic behavior), you could force it with all your might at once, then that would be linear or perhaps exponential (depending on your strength etc..)-the types go on and on.... That's about as simple an example I can think of.
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