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Coupled mass problem with orthogonal oscillations

  1. Mar 16, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider a light elastic string of unstretched length ##4a_o##, stretched horizontally on a smooth surface between two fixed points a distance ##4a## apart. (##a > a_o)##. Three particles of mass m are attached so as to divide the string into four equal sections. Number the segments from left to right ##i = 1 - 4##. The tension ##T_i## in each segment ##i## is proportional to its extension ##(a-a_o)##, with the elastic constant being c>0.

    Suppose that the particles are constrained such that they are able to move only perpendicular to the line along which the string is connected. The system as a whole is planar.

    1) Write down the eqns of motion for the vertical displacement ##x_i## under the assumption that displacements are small. Keep only linear terms in ##x_i/a##. Show that in this approximation, the eqn takes the form ##\underline{\ddot{x}} + n^2 A \underline{x} = 0## and determine ##n## and the matrix ##A##.

    3. The attempt at a solution
    (See diagram for picture)
    For mass m1, it is acted upon by T1 and T2. Defining two angles, ##\theta, \alpha## I get $$m_1 \ddot{x_1} = T_1 \sin \theta + T_2 \sin \alpha \Rightarrow m_1 \ddot{x_1} = T_1 \theta + T_2 \alpha$$ in the small angle approx.

    To get expressions for the acceleration of the other masses, I defined another two angles ##\beta, \gamma## but the expressions were the same, when I sub in the below for the derived tension force.

    Since the system is coupled, if mass m1 moves up then m2 also moves up. So change in length will be ##x_i - x_{i-1}##. This gives ##T_i X = c[a - a_0 + x_i - x_{i-1}]X, X## either ##\theta, \alpha, \beta, \gamma##. When I sub these in, rearrange I don't get the required form that I need to get to. I did a longitudinal version of this problem , and so I am trying to extrapolate what I did in that problem into this problem. Since we are dealing only with a perpendicular oscillations, the problem is rather similar. I think the ##x_i/a## terms come from expressing sin(X) = xi/a

    I can post my working in pen showing detailed drawings etc if required /easier to understand my workings.

    Many thanks.
     

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    Last edited: Mar 16, 2013
  2. jcsd
  3. Mar 16, 2013 #2
    You are making the same mistake you did on another coupled problem here recently. Don't assume any simple geometric relationship between the positions of two masses. Treat their coordinates independently. Their only relationship is via the force of tension.
     
  4. Mar 16, 2013 #3

    CAF123

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    In the horizontal oscillations problem the correct tension force was ##T_i = c( a - a_o + x_i - x_{i -1})##. The set up of that problem was exactly the same, except there we had movement of masses longitudinal and not transverse. Is there something i can do with that to apply it to this problem? Do I need to express the tension force in terms of the displacements x1, x2 and x3?
     
  5. Mar 17, 2013 #4
    You just need to find the forces of tension. You started (almost) correctly: the force acting on the first mass is ## T_1 \sin \theta + T_2 \sin \alpha ## Now, ## T_1 = c (\sqrt {a^2 + x_1^2} - a_0) ##, and ## \sin \theta = \frac {x_1} {\sqrt {a^2 + x_1^2}} ##, so the first term of force is ## c (x_1 - \frac {a_0 x_1} {\sqrt {a^2 + x_1^2}}) \approx c(x_1 - \frac {a_0}{a} x_1) = c(1 - \frac {a_0}{a}) x_1 ##. The second term looks trickier, but observe that the geometry involved is the same, provided we replace ## x_1 \rightarrow x_1 - x_2 ##, so we can fast forward to ## c(1 - \frac {a_0}{a}) (x_1 - x_2)##, and the force acting the first mass is ## c(1 - \frac {a_0}{a}) (2x_1 - x_2)##.

    Now let's go back to the "almost" remark above. Think about the direction of the force and the sign you should assign to ## c(1 - \frac {a_0}{a}) (2x_1 - x_2)##.
     
  6. Mar 17, 2013 #5

    CAF123

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    Hello voko,
    Could you explain in a little more detail how you arrived at this expression for ##T_1##? I reliase the term you have computed is the hypotenuse of the triangle and so you are saying that this is the extension from the natural length of the string.

    EDIT: Also, does the question not say the tension is proportional to a - a0?


    I am not sure I really understand this step.

    If I take y as positive down, then ##T_1 \sin \theta## acts downwards and so would be postive.
     
    Last edited: Mar 17, 2013
  7. Mar 17, 2013 #6
    Yes, this is the hypotenuse (h) and so is the length of the string between its left-most point and the first mass. Then the force is ## c(h - a_0) ##.

    The second triangle is very much alike the first triangle. Except its vertical cathetus is ## x_1 - x_2 ##.


    What is "y"?

    Note it is not important whether "up" is positive or negative. What is important is the direction of ##x_1## and the direction of the force acting on the first mass. Assuming ##x_2 = 0 ##, is it the same direction?
     
  8. Mar 17, 2013 #7

    CAF123

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    It's still not clear, sorry. What second triangle is alike the first and why did you let ##x_1 \rightarrow x_1 - x_2## in the first place? What does that arrow mean? A mapping?

    Define coordinate system with the positive y direction downwards.

    So if m2 stays in place (x2 = 0) and m1 is displaced by x1 vertically (is this what you meant?) then I think the force (T1 sin θ) would still point downwards. If x2 = 0 and x1 = 0 then T1 points leftwards and T2 rightwards.
     
  9. Mar 17, 2013 #8
    Refer to your drawing. The first triangle is the left-most triangle there. The second triangle is the one that follows it, whose vertices are m1 and m2 and the unmarked point under m1 and to the left of m2 (so that the triangle is right).

    The arrow means "replace with", just like the text says.

    Alternatively, you could consider the case of a string which is stretched to a horizontally, and then its ends are moved vertically so that the vertical distance between them is d. Find out the vertical force of tension in terms of a and d. Then you could translate this to any segment in the original problem, by expressing d via x's.

    If x2 = 0, the force is downward if x1 is upward. What does that mean with regard to its sign?
     
  10. Mar 17, 2013 #9

    CAF123

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    This means the force is a restoring force so it will have a negative sign. (If upwards is positive) If I define downwards as positive, then the displacement upwards will be negative then the force will be positive. Since the choice does not matter (as you said), take upwards as postive and so the force will be negative.
     
  11. Mar 17, 2013 #10
    Which makes the first equation ## m\ddot{x_1} = -c(1 - \frac {a_0} {a})(2x_1 - x_2) ##.
     
  12. Mar 17, 2013 #11

    CAF123

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    Indeed. Before I continue with finding expressions for m2,m3 in terms of x1,x2.. I want to understand your first post well.

    So for the first mass, I can write: $$m\ddot{x_1} = -T_1 \sin \theta + T_2 \sin \alpha = -(c(\sqrt{x_1^2 + a^2} - a_o) \cdot \frac{x_1}{\sqrt{x_1^2 + a^2}}) + c(\sqrt{x_2^2 + a^2} -a_o) \cdot \frac{x_2}{\sqrt{x_2^2 + a^2}}. $$Taking the small displacement approx gives in the end, $$c(-x_1 - \frac{a_0}{a}x_1 + x_2 - \frac{a_o}{a}x_2)$$
    I don't think it is the same as what you got?
     
  13. Mar 17, 2013 #12
    No it is not. Specifically, the second term is wrong. I advise that you do the "alternative" problem in my post #8. Then apply the solution to the second term.
     
  14. Mar 17, 2013 #13

    CAF123

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    So a string has some extension a in the horizontal direction. It then has some vertical extension d. So the distance between the ends is ##\sqrt{d^2 + a^2} ##. So ##T_1 = -c(\sqrt{x_1^2 + a^2} - a_o)## since ##x_1## is the vertical distance in this case.

    For m2, the distance between the ends are x1 + x2. So then ##T_2 = c(\sqrt{(x_1 + x_2)^2 + a^2} -a_o)##
     
  15. Mar 17, 2013 #14
    This is correct, but this is not the end. Work out the linear formula for the isolated case, then use it for all the segments.
     
  16. Mar 17, 2013 #15

    CAF123

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    Ok, so in general $$T_i = c(\sqrt{(x_j + x_i)^2 + a^2} - a_o)$$, ## j = i - 1.##

    When I apply this to m2, I get:
    m2 moves downwards, so the force points upwards i.e positive in the defined coord. system.
    So, $$T_2 \sin \alpha = c(\sqrt{(x_1 + x_2)^2 + a^2} -a_o) \cdot \frac{x_1 + x_2}{\sqrt{(x_1 + x_2)^2 + a^2}}$$
    When added to the expression for T1sinθ, I get $$-cx_1 -ca_ox_1/a + c(x_1 + x_2) - \frac{ca_o(x_1 + x_2)}{\sqrt{(x_1+x_2)^2 +a^2}} = + cx_2 - 2ca_ox_1/a - cx_2 a_o/a = -c(2a_ox_1/a - x_2 + x_2 a_o/a)$$ which is close.
     
  17. Mar 17, 2013 #16
    I just noted that you let d = x1 + x2 in #13. This cannot be correct. If x1 = x2, then you get a positive vertical distance, while it should be zero.

    Secondly, you did not get the linear formula of force in terms of d. You got Ti (assuming you correct the x1 + x2 issue), but what you need is its vertical component.
     
  18. Mar 17, 2013 #17

    CAF123

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    If m1 goes up by an amount x1 and m2 down by an amount x2, then the distance between them is x1 + x2. Since the string is also stretched in horizontal direction, I sum the squares of these and sqrt (to get the distance between the two ends of that piece of string)?

    So, like $$|T_i| = c(\sqrt{(x_i + (?) x_j)^2 + a^2} -a_o) \cdot \frac{(x_i + (?) x_j)}{\sqrt{(x_i + x_j)^2 + a^2}}$$ (I added a question mark because still unsure about this)
     
  19. Mar 17, 2013 #18
    You can't use different sign conventions for x1 and x2. If m1 goes up, x1 is positive; if m2 goes down, x2 is negative.

    I suggest you forget about x_i and x_j for a second and solve the simpler problem with vertical separation d - down to expressing the force linearly in d. Then you can substitute x_i and x_j.
     
  20. Mar 17, 2013 #19

    CAF123

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    Then I have d = x1 - x2 for m2


    $$|T_i| = c(\sqrt{d^2 + a^2} - a_o) \cdot \left(\frac{d}{\sqrt{d^2 + a^2}}\right)$$
    EDIT: Forgot to express down to linear in d...

    $$= c(d) - ca_o(d)/a$$
     
  21. Mar 17, 2013 #20
    Linearize this.
     
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