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    Gosh... been sitting here for about an hour trying to sort this out

    "A particle of mass 2kg is attached to the mid-point of a light elastic string of NL 1.5 m and modulus 20N, where A is the ceiling. The other end of the string (B) is attached to the floor which is 3m below the ceiling. Given that P is where particle is in equilibrium"

    a) calculate the distance AP
    b) prove SHM
    c) amplitude and time period

    I can't get the first bit done. I tried using T = mg, with the standard formula but I can't seem to get it to work. I tried loads of different extensions but didn't seem to get the right one.

    For the second bit my omega doesn't seem to be correct, I used the forumla where either side of the string is considered to be separate and identical and use F=ma to sort that out. Didn't work.

    Any ideas?
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    For part a) The string is stretched overall to 3m{extension of 1.5m}, with the particle in the middle initially, but in equilibrium in stretches the top half of the string by Xmeters and then shortens the bottom half of the string by Xm. It is also necessary to know that the bottom half will have some tension and not completely slack.

    Hence you can write an equation;
    Spoiler:
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    T1 = T2 + mg ......where T1=tension in top half of the string and T2= tension in the bottom half of the string.
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    (Original post by bruceleej)
    For part a) The string is stretched overall to 3m{extension of 1.5m}, with the particle in the middle initially, but in equilibrium in stretches the top half of the string by Xmeters and then shortens the bottom half of the string by Xm. It is also necessary to know that the bottom half will have some tension and not completely slack.

    Hence you can write an equation;
    Spoiler:
    Show
    T1 = T2 + mg ......where T1=tension in top half of the string and T2= tension in the bottom half of the string.
    I tried that already. I don't think I put the right extensions down.
    For the T1 i put the extension as 3 - x - 1.5. I have a feeling that is wrong. For T 2 i put 3 + x - 1.5.
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    For T1, the extension is (x+1.5), with Nat.Length=0.75m and for T2 the extension is (1.5-x).
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    Parts c) and cannot be done, with the information which you have provided. I can only do part b) If I assume that the particle is pulled down by by a further x meters below the equilibrium position.
    Are you sure youve given the full question?
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    (Original post by bruceleej)
    Parts c) and cannot be done, with the information which you have provided. I can only do part b) If I assume that the particle is pulled down by by a further x meters below the equilibrium position.
    Are you sure youve given the full question?
    Ooops.. The particle is further pulled 0.1m.
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    (Original post by Shrayans)
    Ooops.. The particle is further pulled 0.1m.
    Yep, I think thats all I need...Managed to work out part a) yet?
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    (Original post by bruceleej)
    Yep, I think thats all I need...Managed to work out part a) yet?
    Yes
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    It works

    Thanks alot!! Reps!!
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    Part b) is similar to a)
    Your value for  \omega^2 should be;

    Spoiler:
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     \omega^2 \,=\, \frac{80}{3}


    P.S: you are very much welcome.
 
 
 
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Updated: March 10, 2008

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