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Mass connected with three springs-Oscillations

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m is attached to three springs A , B , C of equal force constants k as shown in the figure.The angle between each pair of springs is 120° initially . If the particle is pushed slightly against the spring C and released, find the time period of oscillation.

    2. Relevant equations



    3. The attempt at a solution

    Let the particle be displaced from O to P .i.e OP=x . Spring C is compressed by distance x .Spring A is stretched by a distance PA - OA . Spring B is stretched by a distance PB - OB.
    Perpendicular OX and OY are dropped on PA and PB respectively . For small displacement BY ≈ BO and AO ≈ AX .So, PX is approximately extension in spring A .PX=POsinα or PX=POcosβ .

    Now , how do I determine the angle α or β so that I may proceed with the problem ?
     

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    Last edited: Dec 17, 2012
  2. jcsd
  3. Dec 17, 2012 #2

    ehild

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    The distance PO is given, say d. You can find PA with the Law of Cosines. (The angle POA is 120°. )

    ehild
     
  4. Dec 17, 2012 #3
    Let the length of spring be L
    OA≈XA =L

    If we consider ΔOPA then [itex]L^2 = d^2 + (PX + L)^2 - 2d(PX+L)Cosβ[/itex]

    PX is the stretching in spring which is ultimately what we are trying to find and is an unknown and so is L .How will we get a definite value of β ? β should turn out to be 60° .
     
  5. Dec 17, 2012 #4
    There is no need of introducing β. Use Law of Cosines as ehild pointed out.
     
  6. Dec 17, 2012 #5

    ehild

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    You need to find the resultant force acting on the point mass which is displaced towards C, say, with distance d. The spring C exerts force of magnitude kd, the other strings exert forces of magnitude k(L'-L) where L' is the stretched length of the springs connected to A and B. L' can we obtained with the Cosine Law.
    These forces act along the springs, and you need the component parallel with spring C which is k(L'-L)cos(β). β is very close to 60°.

    ehild
     
  7. Dec 17, 2012 #6
    ehild...thank you very much :smile:
     
    Last edited: Dec 17, 2012
  8. Dec 17, 2012 #7

    ehild

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    Have you solved the problem? What is the result?

    ehild
     
  9. Dec 17, 2012 #8
    Force due to spring C=kd
    Component of Force due to spring A =k(xsin60°)(cos60°)=kd/4
    Component of Force due to spring A =k(xsin60°)(cos60°)=kd/4

    Total restoring force =(3/2)kd

    Effective force constant = (3/2)k

    Time period = 2∏(√(2m/3k)) which is the correct answer .

    The problem for me was to get the stretching in the springs . Though I still havent been able to apply Law of Cosines .I found the stretching from isosceles triangle OXA .Angles opposite to equal sides must be equal , hence ∠AOX=90° .Now∠AOP=120° .So ∠POX =30° .

    If I apple Law of cosines in ΔAOP

    L'2 = L2 + d2 -2(d)(L)cos120°

    L'2 = L2 + d2 +(d)(L)

    Now I am stuck here ?
     
  10. Dec 18, 2012 #9

    ehild

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    What you have shown in post #1 is not an isosceles triangle if the angle is 90°, or if it is isosceles triangle the angle is not 90°. But you are right , the angles at the base of a very elongated isosceles triangle can be approximated by 90°. It is a usual approximation, and the solution is correct.
    What I suggested, it would have been a more rigorous derivation, applying calculation with small quantities. That means keeping only the linear terms of a small quantity, and using some approximate formulas, useful to remember:

    1/(1+δ) a≈1-δ, √(1+δ)≈1+δ/2, sin(δ)≈δ, cos(δ)≈1.

    (They follow from the Taylor expansion, stopping at the linear term.)



    L'= L2(1 + d2/L2 +d/L)

    δ=d/L <<1, its second power can be ignored, and applying √(1+δ)=1+δ/2, L'=L(1+d/(2L))

    The stretching of the spring is L'-L=d/2. It should be multiplied by cos(β. You can show that β=pi/3-α. sin(α) is some multiple of d/L, a small quantity, cos(α)≈1, so cos(β)≈1/2+√3/2sin(α).
    It is multiplied by d/L,

    d/(2L)(1/2+√3/2sin(α))=d/(4L) +a second-order small quantity, ≈d/4L.

    ehild
     
    Last edited: Dec 18, 2012
  11. Dec 18, 2012 #10
    Thanks for the wonderful explaination :smile: It had never crossed my mind there was so much going behind the scenes . :biggrin:

    You have been kind enough to explain the maths .

    I am trying to learn some maths here , so was unable to comprehend certain things...

    How?

    How did we get the second term ? Shouldnt it be negative ?

    Why are we multiplying with d/L ?
     
    Last edited: Dec 18, 2012
  12. Dec 18, 2012 #11

    ehild

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    β=π/3-α . See picture. You can understand that alpha decreases with d. If d is very small with respect to L, so is alpha (with respect to pi)
    You need the component of k(L'-L), which is cos(β)k(L'-L).

    cos(pi/3-α)=cos(π/3)cosα-sin(π/3)sin(-α)=cos(π/3)cosα+sin(π/3)sinα.

    k(L'-L)=kd/2. I made an error here in the previous post.

    So the component of force from string A is k(d/2)cosβ=k(d/2)(1/2+√3/2 sinα) ≈ kd/4 as sinα is a very small quantity.

    ehild
     

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  13. Dec 18, 2012 #12
    Thanks ehild...You have explained the maths behind the problem very nicely :smile:
     
  14. Dec 18, 2012 #13

    TSny

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    Another approach (not necessarily any better).

    See attached figure which shows the three springs in the initial configuration rotated so that spring C is vertical.

    Suppose x is increased by dx. Taking differentials of L2 = x2 + d2 gives 2LdL = 2xdx, so dL = (x/L)dx. Using x/L = 1/2 in the initial configuration yields dL = dx/2.

    So, if spring C is compressed by dx, the other two springs stretch by dx/2.
     

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  15. Dec 18, 2012 #14

    ehild

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    Nice explanation, TSny!

    ehild
     
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