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Simple Harmonic Motion Problem 4

  1. Dec 13, 2012 #1
    Hi friends the problem is -

    https://fbcdn-sphotos-d-a.akamaihd.net/hphotos-ak-prn1/s480x480/155412_2656530589803_1383873256_n.jpg

    Attempt -

    As per the problem states,

    When the compound system will oscillate in its natural frequency, The frequency of the oscillation will be, √[k/(m + M)]

    When m will fall through the height h, it'll gain √2gh speed. For some time let say it u.

    It will impact with the cage. Using conservation of linear momentum in the vertical direction,

    linear momentum before = linear momentum after collision.

    so,

    m.u = (m + M)v

    so v = m.u/(M + m)

    and this position will be the mean position of the oscillations,

    Here the cage + particle system would be at it maximum speed which would be,

    v = m.u/(M + m)

    The maximum speed at the mean position is, v = A.ω

    So, A = v/ω

    Putting the value of v and ω

    I am getting the result like this-

    A = {√[k / (M + m)]} . m√(2gh)/(M + m)

    Which is not the correct answer as per the question states.

    The correct answer of this problem is option (A) as per the question.

    Now I have doubt in the question. Is it correct ? We can not Conserve the K. E. of the system because the collision is not perfectly elastic.

    Please friends help me in solving this Problem.

    Thank you all in advance.
     
  2. jcsd
  3. Dec 14, 2012 #2

    ehild

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    Check your result, you multiplied by ω instead of dividing with it. Otherwise your derivation is correct, and the result given are wrong. The amplitude should depend on both masses.

    ehild
     
  4. Dec 14, 2012 #3

    TSny

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    Is this true? Where would the system hang at rest with both masses M + m?

    Need to reconsider this if the mean position of the oscillation is not the collision point.

    Now maybe I'm not seeing something. I don't get anything close to any of the answers given.
     
  5. Dec 14, 2012 #4

    AlephZero

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    There is a sneaky way to do questions like this. You don't have find the right answer yourself. You just have to select which of the given answers is right :smile:

    Look at the four answers and ask yourself
    - Which of the fornulas gives a length? The formulas that don't give a length must be wrong.
    - Of the formulas that are left, which ones don't contain some variable that will obviously affect the answer? (Don't do the math, just use common sense). Those formulas are also wrong.
     
  6. Dec 14, 2012 #5

    ehild

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    You are right, the initial position of the cage ΔL0=Mg/k is not same as the mean position of oscillation ΔL=(M+m)g/k.

    Taking downward positive, the SHM y=Asin(ωt+θ) starts with y0=-mg/k and v0=um/(m+M).
    Anyway, the given answers can not be correct.

    ehild
     
  7. Dec 14, 2012 #6

    Well the first statement is good. I am getting answer with its help. But maths is not allowing this. Now I am thinking that, I have taken the mean position in wrong way. So all the calculation is wrong. But the formulas I used, I think I am correct in that.

    If I leave this technique also(i.e using the mean position and the maximum speed during SHM), applying energy conservation is also not providing the correct answer.
     
  8. Dec 15, 2012 #7

    haruspex

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    You appear to have multiplied by ω instead of dividing by it.
     
  9. Dec 15, 2012 #8
    haruspex:

    The general equation of SHM is, x = A sinωt.

    SO the velocity of the particle would be,

    v = Aω cosωt

    This velocity will be maximum when cosωt will be maximum and ±1

    so Vmax = ± Aω

    hence, A = v/ω
     
  10. Dec 15, 2012 #9

    haruspex

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    I have no problem with any of that. It was the next line that I quoted, where you seem to have done A = ωv instead.
     
  11. Dec 15, 2012 #10

    SammyS

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    It seems to me that there is another factor to be considered. That is, during the time in which the body (of mass m) is in free fall, the cage begins an upward acceleration (the beginning of SHM motion, with ω' = √(k/M) ) since the cage-spring system will not be in equilibrium.
     
  12. Dec 15, 2012 #11

    TSny

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    Oh, interesting! So, we are to assume the mass m was initially attached to the cage before it fell. Back to the drawing board for me.
     
  13. Dec 15, 2012 #12

    TSny

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    Thanks, SammyS. I believe it does indeed work out to be answer (A). [EDIT: Dang. Found an error. Not getting a simple answer.]
     
    Last edited: Dec 15, 2012
  14. Dec 15, 2012 #13

    haruspex

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    I see nothing in the problem statement that implies the cage-spring was not in equilibrium initially. Indeed, if you don't assume that there's not enough info to solve the problem.

    EDIT: Ok, I see what you're suggesting - that the mass m was initially on some support within the cage, then fell from it. If that's correct,m the problem is very poorly worded, but I admit that with the more obvious reading I get none of the offered amplitudes. (I get the rebarbative √{(2mg/k)(mh/(M+m) + (M+m)g/k)}.)
     
    Last edited: Dec 15, 2012
  15. Dec 15, 2012 #14

    SammyS

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    Well, I haven't worked it all the way through myself, but I suspect the answer should depend on both M and m , so answer (A) looks suspect. (Other answers ruled out per AlephZero)
     
  16. Dec 15, 2012 #15

    SammyS

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    I agree.

    I am referring to the time during which the body, of mass m, is in free fall. I assume that before the body is in free fall, the cage-body-spring system is in equilibrium.
     
  17. Dec 15, 2012 #16

    TSny

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    I think SammyS is interpreting the problem as saying that the mass m was initially supported inside the cage at a height h above the bottom of the cage (maybe hanging from a string attached to the top of the cage). Everything is initially at rest in equilibrium when suddenly the string supporting m breaks.
     
  18. Dec 15, 2012 #17

    haruspex

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    Yes, just realised that. I edited my earlier post.
    But I don't think it's going to lead to a solution. We have to compute the KE remaining after impact. To get that, we need the collision speed, and that appears to involve solving a mixture of functions equation.
     
    Last edited: Dec 15, 2012
  19. Dec 15, 2012 #18

    TSny

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    For the record, I get almost the same result for the original interpretation. Only difference is that I have m instead of (M+m) in the last term.
     
  20. Dec 15, 2012 #19
    OMG. Sorry Really I did mistake out there
     
  21. Dec 15, 2012 #20
    Well sorry to interrupt but my motive is to solve the problem. And I think any assumption couldn't work there SammyS.
     
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