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Spring and string pendulum (oscillations)

  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data

    Hi everyone! Here is a new problem about oscillations! Thx to all of you, I'm definitely making progress in the field. Let's see how that problem goes:

    A pendulum of mass m is hanging on a string of length L and is "pushed" by a spring with spring constant k. At the deepest point of the pendulum (that is, when α = 0), the spring is compressed at x0 of its equilibrium position.
    a) determine the equation of motion of the angle α (for the case α<<1).
    b) the solution of the equation is α(t) = A⋅sin(ω⋅t) + B⋅cos(ω⋅t) + αR. Find the angular frequency and the angle αR (equilibrium position) in terms of m, L, and k.
    c) what is the solution, when the pendulum goes at t=0 from the position α = 0?

    Check out the two attached picture. The 1st one describes the problem while the 2nd one shows my interpretation of the situation.

    2. Relevant equations

    Equation of motion, oscillations, torque

    3. The attempt at a solution

    a) At (2) (see pic), m⋅x'' = -k⋅x
    ⇔ x'' + (k/m)⋅x = 0

    ΣD = FF ⇒ α'' = -(k⋅x)/(m⋅L)

    For a small angle we can say that (x/L) ≈ α, so the equation of motion goes:

    α'' + (k/m)⋅α = 0

    I'm pretty confident that it is correct, but let's see about b) :

    I can input the solution to the equation in my equation for t=0, and after simplifying the cos and sin I get:

    -B⋅ω2 + (k/m)⋅B + (k/m)⋅αR = 0

    I do the same for α(0) and I find that αR = 2⋅B. I input that in the previous equation and get, quite easily:

    ω = √(3k/m)

    Now I've tried a few things, mixing equations and states, but I can't get my hands on αR! Could someone give me a clue in what direction I should look?

    For c) I'm not sure I understand the question right. Isn't it what I have already done in B on the way of solving for ω?


    Thank you very much in advance for your advices, I appreciate it!


    Julien.
     

    Attached Files:

  2. jcsd
  3. Apr 1, 2016 #2
    somehow i present a different picture- the initial condition given shows that the pendulum had come from some angular displacement and spent its entire energy in pushing the spring to a length x0 and now the spring will unwind and push the bob therefore the oscillation will continue with again the bob coming back and pushing the spring .
    well in this situation at any instant of time t after the event if the displacement of the bob is alpha the two forces will act
    1. the push by spring 2. the usual mgsin(alpha) and the equation of motion in angular coordinate will have these two terms. then its solution can be found.
    if i am correct the new eq. will look different a term g/l will get added to -k/m term.
     
  4. Apr 1, 2016 #3
    @drvrm Mm that's interesting, I thought t=0 was when α=0 (that is, when whoever pressing the pendulum again the spring releases it into motion). I'll give some thoughts to your interpretation.
     
  5. Apr 1, 2016 #4
    your problem statement does not identify who is pushing and releasing it.
    even an agency is there again at any angular displacement the mgsin alpha term will get added.
     
  6. Apr 1, 2016 #5

    haruspex

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    I disagree. We are not told the initial state of motion.
    At first, we are only told that when the string is vertical the spring is compressed x0 from the equilibrium position (note, not 'to' a length x0). We are not told a time or velocity corresponding to that position.
    In part c, we are told to take it as being in the vertical position at t=0, but we still do not know the velocity at that time, so do not know the amplitude.
    But I agree that gravity needs to be taken into account.
     
  7. Apr 1, 2016 #6
    Indeed otherwise the pendulum would not come back down right? I get back to it.
     
  8. Apr 2, 2016 #7
    Okay I've attached my new interpretation of the situation (it's actually very similar apart from dealing with g) and my calculations for a). I now get this equation of motion:

    α'' - ((g/L) + (k/m))⋅α = 0

    Is that correct? Then for b) I input the solution into the equation of motion and get:

    -A⋅ω2⋅sin(ω⋅t) - B⋅ω2⋅cos(ω⋅t) - ((g/L) + (k/m))(A⋅cos(ω⋅t) + B⋅sin(ω⋅t) + αR) = 0

    I check out what happens for t=0 (is that allowed?? Is that the right way to proceed?) :

    -B⋅ω2 - ((g/L) + (k/m))(B + αR) = 0
    ω2 = -((g/L) + (k/m))(1 + αR/B)

    I'm blocked here at the moment. I tried to find an expression for B in terms of αR in the first and second derivatives of α(t), but I've been unsuccessful so far.

    Julien.
     

    Attached Files:

  9. Apr 2, 2016 #8

    haruspex

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    Your suspicions should be aroused immediately. According to that equation, the more alpha is positive the more its acceleration will be positive, leading to a runaway acceleration, not SHM. Check the signs.
     
  10. Apr 2, 2016 #9
    @haruspex I am really struggling. But those mistakes do help me to understand better what's going on there.
    I have redone it and indeed get α'' + ((k/m) + (g/l))⋅α = 0. I input the solution in there and get:

    A⋅sin(ω⋅t)⋅(ω2 - ω02) + B⋅cos(ω⋅t)⋅(ω2 - ω02) - ω02⋅αR = 0
    with ω0 = √((k/m) + (g/l))

    Then I rewrote the expression with ω = ω/ω0 as I have observed some people do and I reach:

    A⋅sin(ω⋅t)⋅(ω2 - 1) + B⋅cos(ω⋅t)⋅(ω2 - 1) - αR = 0

    If I set t = 0, this simplifies to:

    ω2 = 1 + αR/B

    Now (if that is -for once- correct), the ideal would be an equation with B and αR allowing me to replace one or the other. αR is only present in α(t), and for A to disappear t must be equal to zero. The only problem is that I don't know at which angle α the pendulum is going to be at that precise moment. Maybe there's another way.

    I'm still working on the problem, but I wanted to update where I am standing now. Is that better, or is it still a disaster? :DD Thank you very much again!


    Julien.
     
  11. Apr 2, 2016 #10
    I just had an idea to solve for ω. If I set up α(t) = 0, that means the pendulum is at its deepest point against the spring. Because it is the point where it changes direction, its velocity must also be 0, allowing me to use α'(t) = 0 as well. Is that valid?


    Julien.
     
  12. Apr 2, 2016 #11

    haruspex

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    No, when it says the deepest point it just means the lowest point in the swing. It does mot mean this is the extreme end of the swing.
    I don't understand why you have both ω and ω0. Your expression for ω0 is the frequency, no?
    In your equation
    there is a term missing. Remember, the equilibrium position is not at alpha=0.
    Suppose at the equilibrium position (αR) the spring is compressed by x'. Write an equation about that.
    When α=0, it is compressed a further x0. Express the spring compression when at angle α in terms of L, α, x' and x0.
     
  13. Apr 3, 2016 #12
    @haruspex Thank you for your patience, I don't know how you can stand all my mistakes. :oops: Okay now what about that:

    At the equilibrium position the horizontal forces balances each other, thus m⋅g⋅sinα = -k⋅x'.

    When α=0, only the spring force -k⋅(x' + x0) acts horizontally, and I can derive from the torque the angular acceleration:

    α'' = k⋅(x' + x0)/(mL) = -(g/L)⋅sinα + (k/m)(x0/L)
    = -(g/L)⋅α - (k/m)⋅αR

    Every time I think I get something right it's wrong, so I'd better wait for a confirmation before going any further :DD

    Julien.
     
  14. Apr 3, 2016 #13

    haruspex

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    Yes, except that should be αR, and you can use the small angle approximation (I assume).
    I don't think we care specifically what the torque is at angle 0.
    What I was looking for there was just a bit of geometry, nothing to do with forces or torques. From that, you can get the spring force in terms of those, and use the equilibrium equation to eliminate x'.
     
  15. Apr 3, 2016 #14
    @haruspex I've attached a picture of how I understood what you said. Hopefully this is correct. I get that -k⋅(x' + x0) = (m⋅g - k⋅L)⋅αR.
     

    Attached Files:

  16. Apr 3, 2016 #15

    haruspex

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    Not bad, but you have some sign errors. Since x is measured from the right, x' etc. are all positive. Mg sin αR is positive too, so there should be no minus sign in that equation. Likewise, in the last equation both minus signs should be plus.
     
  17. Apr 3, 2016 #16
    @haruspex Okay, I've tried to repair the sign mistakes and to derive the equation of motion from that. I get α'' - (g/L)⋅(α + αR) - (k/m)⋅αR = 0. It looks a bit strange, not sure how to interpret it.

    I've attached my calculation steps.

    Julien.
     

    Attached Files:

  18. Apr 3, 2016 #17

    haruspex

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    A couple of errors in your first ##\ddot \alpha## equation. As alpha increases, the gravitational force should act to reduce alpha, and the spring's tendency to increase alpha should reduce. So again some signs wrong.
    Also, I don't understand the presence of x0. This equation should be for a general position x, not the vertical position.
     
  19. Apr 3, 2016 #18
    @haruspex For the x0 I was thinking x all along, I replaced it now. About the signs, I now get for the torque ∑τ = -k(x' + x)⋅L - m⋅g⋅sinα⋅L. I've attached the equation of motion.

    Julien.
     

    Attached Files:

  20. Apr 3, 2016 #19
    I find dealing with the signs very confusing. When I will be finished with this problem (and that is apparently going to take a while), I'll post the whole solution again for people searching infos about the topic.
     
  21. Apr 3, 2016 #20

    haruspex

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    x is defined as the total compression. No need to add x0 to that.
    Also, you did not need to switch the sign on that term. Sorry if I gave you the impression you should. The spring force acts to the right, so tends to increase alpha. It's just that at larger alpha, that tendency is less.
     
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