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Pendulum and spring; Would like to check if the result is correct.

  1. Mar 12, 2016 #1
    1. The problem statement, all variables and given/known data
    Spring connected to a pendulum holding a ballwith mass m, length l, spring constant k, spring deflection x, angle of pendulum alpha, angular velocity w.


    2. Relevant equations
    Derive the angular acceleration

    3. The attempt at a solution

    I made the three body diagram, and find that -k*x-m*g*sin(alpha)=m*a, from here we know that a=angularacceleration*l, x=l*sin(alpha),alpha=w*t and w=sqrt(k/m+g/l) ---> not sure about this one. And I obtained that angularacceleration=w^2*sin(w*t). Is my derivation correct?
     
  2. jcsd
  3. Mar 12, 2016 #2

    haruspex

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    Not sure I have the picture. Is there a simple pendulum with a fixed length rod, and a spring connected to the same mass?
    If so, how is the spring arranged... vertically below, horizontal...?
    Or is the 'rod' itself the spring? Or .... .?
     
  4. Mar 12, 2016 #3
    Is a pendulum with a fixed length rod and a spring connected to the same mass, in horizontal position. I will try to picture it.
    \
    \
    \
    ><><><><● where: \ rod, ● mass and >< spring
     
  5. Mar 12, 2016 #4
    As I can see the result of my "picture" is not the one I was expecting, assume that the rod goes to the ball, and you will see it clear
     
  6. Mar 12, 2016 #5

    haruspex

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    Ok. Are we to take the spring as sufficiently long compared with vertical displacements of the mass that the spring is effectively horizontal always?
     
  7. Mar 12, 2016 #6
    Yes, it is always effective
     
  8. Mar 12, 2016 #7

    haruspex

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    I guess you mean always horizontal.
    You seem to have used w for two different entities, angular velocity and frequency.
    Your answer looks wrong since k does not feature.
    Try to write the complete differential equation for the horizontal motion, either in terms of spring displacement, x, or in terms of angular displacement, alpha.
     
  9. Mar 12, 2016 #8
    Here you have the differential eqution, my problem is if in this case the W is sqrt (g/l+k/m)
     

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  10. Mar 12, 2016 #9

    haruspex

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    Yes, that looks right for the frequency, but as I wrote, you have w standing for two different things, frequency and angular velocity.
    Please post your working for your final equation for angular acceleration.

    Are you sure you have stated the question correctly, that you are to find the angular acceleration as a function of the angular velocity?
    I don't see how you can do that without knowing the amplitude.
     
  11. Mar 12, 2016 #10
    yes, you are right, i made a mistake while saying that w was angular velocity, w is the frequency. Therefore we will obtain that acceleration(angular)=-w^2*sin(alpha), now alpha would be substituted by... angular velocity*time?
     
  12. Mar 12, 2016 #11

    haruspex

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    If the SHM is given by ##\theta=\Theta\sin(\omega t)## then what are the angular velocity and angular accelerations as functions of t?
    How can you use that to write the angular acceleration as a function of angular velocity?
     
  13. Mar 13, 2016 #12
    the angular velocity is Θ*w*cos(w*t) and the angular acceleration is -Θ*w^2*sin(w*t). I think alpha should be substituted by alpha initial + w*t, this is the only solution I can think about
     
    Last edited: Mar 13, 2016
  14. Mar 13, 2016 #13

    haruspex

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    There is still some confusion here we need to clear up. The problem statement reads:
    ?
    It doesn't make any sense to me to change that to "frequency w" since you can and have found the frequency. If you want to use w for frequency then we need to assign another symbol for angular velocity. Call it z.
    Yes, you can substitute alpha for the wt. We effectively fixed alpha initial by not putting a phase term in the equation. When alpha is zero the acceleration is zero, so alpha initial is zero.

    Edit: that was wrong! alpha is not the same as wt. they are quite different beasts. See later post.
     
    Last edited: Mar 13, 2016
  15. Mar 13, 2016 #14
    Definetely the mistake is mine. we do not have the angular velocity, but the "frequency w". Therefore, I think we arrived to the solution we were looking for: -w^2*sin(w*t+alpha), where we know the angle, and the frequency w. Is that correct?
     
  16. Mar 13, 2016 #15

    haruspex

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    No, I'm sorry, I was wrong to endorse substituting alpha for wt. (Post edited.)
    wt is not an actual angle, it is a theoretical angle representing where the system is in its oscillation.
    In the generic SHM equation I quoted in post #11, ##\theta=\Theta \sin(\omega t)##, where does alpha fit?
     
  17. Mar 13, 2016 #16
    I think in our case alpha should be equal to Θsin(ωt), as this formula is telling us what is the position of the system for every time unit.
     
  18. Mar 13, 2016 #17

    haruspex

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    Right. So can you use that to express the angular acceleration as a function of alpha and w?
     
  19. Mar 13, 2016 #18
    yes, the final result would be -w^2*sin(Θsin(ωt)) where Θ would be the elongation of the system
     
  20. Mar 13, 2016 #19

    haruspex

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    No. How do you get that?
     
  21. Mar 13, 2016 #20
    substitution of alpha by Θsin(ωt), is not what we said it was correct in post #17?
     
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