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Suddenly removing a spring

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    A mass is between 2 springs between 2 walls. The mass oscillates with amplitude d. The springs are at equilibrium when the mass is in the center. At the moment the mass is in the center a spring is removed. What is the resulting x(t) and the new amplitude? Note the k's are equal.
    Edit: I forgot to add that at t=0 x(t)=d/2 when there is one spring.


    2. Relevant equations



    3. The attempt at a solution
    So I solved for x(t) for both before and after removing the spring.
    Before: x(t) = Asin(2wt) + Bcos(2wt)
    after: x(t) = Asin(wt) + Bsin(wt)
    I'm just assuming the new amplitude is twice what is was since when the spring is removed it has the energy of 2 springs, but now the resistance of one. Is there a way to show this using the x(t)'s I found? I feel like they should relate to the problem more.
     
    Last edited: Oct 2, 2012
  2. jcsd
  3. Oct 2, 2012 #2
    1. With both in equilibrium, means not forces applied to the mass.
    2. If the mass displaced with 2 springs, then with only one spring we have to take conservation of energy into consideration.
     
  4. Oct 2, 2012 #3
    1. It is oscillating with amplitude d, we know forces are being applied.
    2. I know we can use conservation of energy, but is there anyway using the x(t) that i derived?
     
  5. Oct 2, 2012 #4
    Before: x(t) = Asin(2wt) + Bcos(2wt)
    after: x(t) = Asin(wt) + Bsin(wt)
    -------------------------------
    The amplitude must be different for both cases.
     
  6. Oct 2, 2012 #5
    for before I get just x(t) = Asin(2wt) since x(0)=0 but how do I show the amplitude difference with these equations? Edit: Actually scratch that, x(0) = d/2, I'll see where this gets me
     
  7. Oct 2, 2012 #6
    The As and Bs are different in the two equations. You need to express one set via the other. A and B in the initial equation must be expressible via d and the condition that at t = 0 the mass is going through the equilibrium point.
     
  8. Oct 2, 2012 #7
    I can solve for B using x(0) = 0 but for A do i have to use the fact that Asin(wt) + Bcos(wt) should equal 2d at their max, so when wt = pi/4, and then solve for A?
     
  9. Oct 2, 2012 #8
    B (for the original equation) = 0, that's correct. So the original equation is x = A sin 2wt. What is A in terms of d?
     
  10. Oct 2, 2012 #9
    Sorry. I fogot to include an important fact in the question. I just added it. Using that I get B = d/2, so how do I get A? What I was thinking in #7 was wrong, I'm guessing I just find a maximum of my function using derivatives, which won't work either, nevermind.
     
    Last edited: Oct 2, 2012
  11. Oct 2, 2012 #10
    Where does this additional condition come from? Is it part of the problem? What time is t = 0?
     
  12. Oct 2, 2012 #11
    It is from the problem and t = 0 is when the spring disappears (also in the problem). I tried to only type the relevant stuff. I clearly didnt understand the problem fully
     
  13. Oct 2, 2012 #12
    This new condition contradicts the other statement. One spring is said to disappear when the mass is at the center. That means x(0) = 0. x = d/2 means the mass is half-way between the center and the maximum displacement.
     
  14. Oct 2, 2012 #13
    Ok now that makes sense. It does not disappear when m is at the middle, but when x=d/2 at which t=0. I am really glad I read properly, but I'm still not sure how to find A or the new amplitude
     
  15. Oct 2, 2012 #14
    Does the problem specify in which direction the mass was moving when the spring separated?
     
  16. Oct 2, 2012 #15
    Yes, to the right. How is that relevant?
     
  17. Oct 2, 2012 #16
    To get A, you have to simplify the equation

    [tex] x = A \sin 2 \omega t - B \cos 2 \omega t [/tex]

    Divide it through by ## \sqrt {A^2 + B^2} ##, that will get you

    [tex] x = \sqrt {A^2 + B^2}[ a\sin 2 \omega t - b \cos 2 \omega t ] [/tex]

    where ## a \le 1, \ b \le 1 ## so you could let ##a = \cos \alpha, \ b = \sin \alpha ##, and use trigonometry to transform the equation further.
     
    Last edited: Oct 2, 2012
  18. Oct 2, 2012 #17
    Well don't I already know B=d/2 using x(0), so is there not an easier method?
     
  19. Oct 2, 2012 #18
    Using this method, you convert ## x = A \sin 2 \omega t + B \cos 2 \omega t ## into ## x = \sqrt {A^2 + B^2} \sin (2 \omega t + \alpha) ##, and it is then obvious that ## d = \sqrt {A^2 + B^2} ##.
     
  20. Oct 2, 2012 #19
    I honestly have no idea how or why/how you did that or how you decided that it needed to be done, nor do i see how d is obvious.
     
  21. Oct 2, 2012 #20
    Re how I did that, go the previous page and how another look.

    Re why d is obvious: look at ## x = \sqrt {A^2 + B^2} \sin (2 \omega t + \alpha) ## - what is the maximum value of it? Then recall the definition of amplitude.

    That explains WHY I did that.
     
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