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SHM: mass between 2 prestretched springs

  1. Feb 5, 2008 #1
    1. The problem statement, all variables and given/known data
    This is from Advanced Physics by Adams and Allday. Spread 3.34, Q 3.

    A friction-free trolley of mass 2 kg is tethered to rigid supports at each end by two identical springs of spring constant 20 N/m each. The springs obey Hooke's law and remain in tension as the trolley is displaced 2 cm to one side and released. What is the effective spring constant of the system tethering the trolley?

    2. Relevant equations
    [itex]F = - k x[/itex]

    Where
    • F is restoring force
    • k is spring (stiffness) constant
    • x is displacement from equilibrium position
    3. The attempt at a solution
    In the equilibrium position let each spring be stretched by x generating a restoring force F(each x and F are identical by symmetry).

    Move the trolley [itex]\delta x[/itex] to the right ([itex]\delta x < x[/itex]).

    The extension of the left spring is now [itex]x + \delta x[/itex] and the right spring [itex]x - \delta x[/itex].

    The restoring force from the left spring is now [itex]F_L = - k ( x + \delta x)[/itex]

    The restoring force from the right spring is now [itex]F_R = - k ( x - \delta x)[/itex]

    The net force (on the trolley), acting to restore the previous equilibrium position is
    [itex]F_L - F_R[/itex]
    [itex]= - k ( x + \delta x) - (- k ( x - \delta x)[/)[/itex]
    [itex]= - 2 k \delta x[/itex]

    Thus the spring constant of the system is twice the spring constant of the individual springs, that is 40 N/m.

    The answer given in the textbook is 10 N/m.

    What is the right answer?
     
  2. jcsd
  3. Feb 5, 2008 #2

    Tom Mattson

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    The book's answer is correct. Springs in series add like capacitors in series.
     
  4. Feb 6, 2008 #3
    Thanks Tom but you can't be right. If springs in series add then 20 N/m + 20 N/m = 40 N/m and the book's answer (10 N/m) is not correct.

    Are you referring to extension (in which case they do add) or to their spring constant? The question is about the spring constant.

    This question is not about the usual springs in series with an attachment at one end and load at the other. It is about springs in series with an attachment at each end and load in the middle. Crucially the force of each spring is not the same as it is with the usual springs in series.

    For what it's worth (I don't believe it's relevant) for springs in series, their extensions add but their spring constant is given by:

    [tex]\frac 1 {k_{total}} = \frac 1 {k_1} + \frac 1 k_2 + ... + \frac 1 k_n[/tex]

    Derivation:
    [tex]x_{total} = x_1 + x_2 + ... + x_n[/tex]

    [tex]F = -kx[/tex]

    [tex]x = \frac {-F} {k}[/tex]

    [tex]\frac {-F_{total}} {k_{total}} = \frac {-F_1} {k_1} + \frac {-F_2} {k_2}x_2 + ... + \frac {-F_n} {k_n}[/tex]

    But the same force is applied to each spring (force is a "through variable" as opposed to a "cross variable")

    [tex]F_{total}} = F_1 = F_2 = ... = F_n[/tex]

    [tex]\frac 1 {k_{total}} = \frac 1 {k_1} + \frac 1 k_2 + ... + \frac 1 k_n[/tex]

    See, for example, http://en.wikipedia.org/wiki/Hooke's_law


    Can I rephrase my question as "Is there anything wrong with my calculation (that gives a different answer from the book)?"
     
  5. Feb 6, 2008 #4
    You have a small error in your work, because if the extension of the left spring increases to [itex]x + \delta x[/itex], the compression of the right spring increases to [itex]x + \delta x[/itex] as well.

    The restoring forces [itex]F_L = F_R = -k(x + \delta x)[/itex], so total restoring force is [itex]F_L + F_R = -2k(x + \delta x)[/itex]. Actually, [itex]x + \delta x[/itex] is just the new extension [itex]x_2[/itex] of each of the springs; the restoring force changes as the object moves, as it should, and the spring constant is [itex]2k[/itex].

    I don't know why the book gives 10 N/m as the answer, because intuitively and mathematically it appears to be 40 N/m.
     
  6. Feb 6, 2008 #5
    Thanks Tedjn :)

    I understand the springs remain in tension because the question states "The springs ... remain in tension".

    Wrong answers are not rare in this book! In this case I'm not yet confident enough of my own working to accept that the book's answer is wrong.
     
  7. Feb 7, 2008 #6

    Shooting Star

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    The eqn for the RLC circuit is Ld^q/dt^2 + Rdq/dt + q/C = E.

    The corresponding eqn for a linear oscillator is md^x/dt^2 + bdx/dt + kx = F.

    We know that capacitance decreases when added in series. Since k is in the numerator and C is in the denominator, if C decreases for two capacitors added in series, the k should increase when two springs are added in series, and should be equal to 2k.
     
  8. Feb 7, 2008 #7
    Thanks Shooting Star :)

    If your conclusion is for the usual arrangement diagrammed at http://en.wikipedia.org/wiki/Hooke's_law#Multiple_springs then it is not right. My derivation earlier in this thread and Wikipedia's on the same page show
    [tex]\frac 1 {k_{equivalent}} = \frac 1 {k_1} + \frac 1 {k_2}[/tex]

    For [tex]{k_1} = {k_2} = k[/tex]
    this simplifies to [tex]k_{equivalent} = k / 2[/tex]

    Incidentally, for two springs in parallel:
    [tex]k_{equivalent} = k_1 + k_2[/tex]
    For [tex]{k_1} = {k_2} = k[/tex]
    this simplifies to [tex]k_{equivalent} = 2k[/tex]

    So now we have two questions:
    1. Is my calculation for a mass tethered to rigid supports at each end by two identical springs correct (and the textbook wrong)?
    2. What is wrong with Shooting Star's reasoning?
    I can't answer the second question but do note that
    md^x/dt^2 + bdx/dt + kx = F
    is the equation for forced damped oscillation describing how the externally applied exciting force is distributed amongst accelerating, damping and static elements.

    The original problem is about a natural, undamped oscillation where the applicable equation is
    [tex]m \frac {d^2x} {dt^2} = -kx[/tex]
    this being a special case of the forced damped oscillation with no forcing and no damping.
     
  9. Feb 7, 2008 #8

    Shooting Star

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    The situation is not simillar with the series picture in the link given above.

    There, the extensions and restoring forces are in the same direction and are shared by the two springs. In your case, to come back to grass roots, if one spring is pulling and the other is pushing and the extension or compression is the same in both, the total restoring force has to be twice that of each.

    (I will have to look into my argument again given in post#6.)

    EDIT: This case, with the mass in the middle, is exactly like the case for parallel springs given in the wikipedia link, since both springs are being stretched, as I had mentioned before.
     
    Last edited: Feb 7, 2008
  10. Feb 7, 2008 #9
    Thanks Shooting Star :)

    The original question says both springs remain in tension. Thus, as the trolley moves toward one spring, the tension in that spring is reduced and the tension in the other spring is increased. I believe the total restoring force is still twice that of each.
     
  11. Feb 7, 2008 #10
    It is similar but not exactly like the parallel springs given in the wikipedia link because when the trolley moves one spring is extended more and the other less - the same magnitude but the opposite sense.
     
  12. Feb 7, 2008 #11

    Kurdt

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  13. Feb 7, 2008 #12
    Thanks for finding it and pointing it out. :smile: Similar and equally unusual.

    I suspect the "always in tension" of the original question is immaterial; it doesn't matter if the rest state of the springs is in tension, in compression or neutral as long as they stay linear -- within proportinality, don't sag etc. The rest state determines the absolute forces in a system where only force differentials are effective.

    That would also mean the system, despite my quibble above, is equivalent to a "springs in series" system. Tension (or reduced compression) in a negative direction is equivalent to compression (or reduced tension) in a positive direction.
     
  14. Feb 10, 2008 #13

    Tom Mattson

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    Well, I'll be damned. :redface: I just assumed that the two situations were the same because of transmissibility of forces. Thanks for straightening that out.
     
  15. Feb 11, 2008 #14
    Oh, dear! When I wrote "springs in series" I meant "springs in parallel".

    In my first post I showed that the relationship between restoring force (toward the equilibrium position) and displacement (from the equilibrium position) of this system is equivalent to a parallel system.

    Here it is again, more completely and more generally. Positive is defined as from left to right. Moving the trolley toward the right thus adds to the left spring's extension and equally subtracts from the right spring's extension.

    Subscripts are L for left, R for right and 0 for initial (euilibrium).

    In the equilibrium position, considering forces on the trolley
    [tex]F_{L0} + F_{R0} = 0[/tex]
    [tex]-k_L x_{L0} - k_R x_{R0} = 0[/tex] ([itex]x_{L0}[/itex] and [itex]x_{R0}[/itex] have opposite signs)

    If the trolley is displaced [itex]\delta x[/itex] to the right the restoring force is
    [tex]F_{L0} + F_{R0} = F[/tex]
    [tex]F = -k_L (x_{L0} + \delta x) - k_R (x_{R0} - \delta x)[/tex]
    [tex]= -k_L x_{L0} - k_R x_{R0} - (k_L + k_R) \delta x)[/tex]
    The first two terms are the equilibrium forces from above and sum to 0 so
    [tex]F = - (k_L + k_R) \delta x)[/tex]
    Thus the spring constant of the system is the sum of the spring constants of its component springs, the same as for springs in parallel.

    Nothing in this analysis implies that, in the the equilibrium position, the springs are in tension or compression; if the left spring is in tension it's extension is positive, otherwise it is negative and vice versa.

    Thus the arrangement is exactly equivalent to two springs in parallel.
     
  16. Feb 11, 2008 #15
    I'm sorry; no need for damming; it was my mistake. See my post just before this.
     
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