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Transverse Oscillation of horizontal Mass-Springs system

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data

    Mass M is supported by a smooth table and connected by two light horizontal springs to two fixed blocks. Each spring is of natural length L. Both springs are initially extended by L to get a total width between blocks of 4L. The spring constant of both springs is k.

    When the mass is given a small transverse displacement, y, show that the restoring force is approximately ky.

    See photo below:

    Photo on 11-11-2015 at 15.54.jpg

    2. Relevant equations

    F = kx
    sinθ = o/h


    3. The attempt at a solution

    Restoring force, Fr = 2kxsinθ
    sinθ = y / (2L + x)
    x = (y / sinθ) - 2L
     
  2. jcsd
  3. Nov 11, 2015 #2
    First you should draw a diagram showing a small transverse displacement y of the ball, and draw a coordinate system. Then write your equations. For example, I don't understand why there is an x in your equations at all.
     
  4. Nov 11, 2015 #3
    The x is to do with the extension of the spring which will occur when you displace the ball by y upwards.
     
  5. Nov 11, 2015 #4

    haruspex

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    Are you being consistent about the meaning of x? Is it the total extension of the spring, or the additional extension beyond its extension when y=0?
     
  6. Nov 11, 2015 #5
    Additional extension beyond what it was (L) at y=0.
     
  7. Nov 11, 2015 #6

    haruspex

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    Then reconsider your equation for Fr.
     
  8. Nov 11, 2015 #7
    Ah, good point.

    So it should be: Fr = 2k(L+x)sinθ

    Which would then be: = 2ky(L+x) / (2L+x)

    Agreed?
     
  9. Nov 11, 2015 #8

    haruspex

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    Yes.
     
  10. Nov 11, 2015 #9
    I'm still unsure how to get from there to simply ky however.
     
  11. Nov 11, 2015 #10

    haruspex

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    It says small displacements. If y is small, x is very small. What approximations does that allow you to make?
     
  12. Nov 11, 2015 #11
    Expanding previously, Fr = (2kL+2kx)y / (2L+x), so if x is small then 2kx ≈ 0 and 2L + x ≈ 2L? Which would leave 2kLy/2L = ky but then surely the y on the top would make the whole thing ≈ 0?
     
  13. Nov 11, 2015 #12

    haruspex

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    What we're doing here is finding the first order approximation. You can think of it as writing the full form as the sum of a series of terms in which, as the displacement tends to zero, each term tends to zero faster than the one before it: a power series, say. The first order approximation throws away all except the first term.
     
  14. Nov 11, 2015 #13
    Well, certainly, if you put y = 0, then you don't have to do anything. There are no oscillations! You are making an approximation, where y, and x are small compared to L. So expressions like L + x, where L is large, and x is small, are nearly equal to L. e.g., 1000 + 1 ~ 1000. If you look at the y on top, you cannot neglect it because you don't have anything to compare it with. After all, while 1 << 1000, 1 is not small compared to 0.
     
  15. Nov 11, 2015 #14
    Thank you both for your help, in combination I believe I understand this much better now but I'll continue to ponder this idea of approximations (in terms of power series) over. Much appreciated!
     
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