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Finding Amplitude SHM

  1. Jan 11, 2008 #1
    1. There is a spring connected to a mass on a frictionless horizontal surface. Another mass smaller in size is placed ontop of the bigger mass. It asks find the amplitude so that the smaller mass falls off. Coeffient of friction is 0.4 between the two masses. K is given as 200 N/m




    2. Simple harmonic motion eqn's



    3. F = kx
    friction force between the smaller mass and bigger mass is = mu * Normal
    Normal = smaller mass times 9.81 m/s^2

    so i got my frictional force...

    is that equal to Kx where K is given so i solve for x?
     
  2. jcsd
  3. Jan 11, 2008 #2

    Kurdt

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    The maximum force will occur at the max amplitude of the oscillation so you need to set that equal to the force required to make the block move as you have stated above.
     
  4. Jan 11, 2008 #3
    is this right
    the mass of each blocks are given m = 1.8 kg and M = 10 kg where m is smaller and M is the bigger mass

    the force required to make the small block move:
    normal of the small mass = 1.8 x 9.81 = 17.66 N
    Force required to move small mass = 0.4 x 17.66 = 7.064N

    Now is this the force we are looking to create through the spring

    Im guessing YES

    F= kx

    F = friction force

    therefore

    7.064 = (200)Xm

    Xm = 0.0353 m
     
  5. Jan 11, 2008 #4

    Kurdt

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    Look ok to me.
     
  6. Jan 11, 2008 #5
    ok THANKS A LOT BUT................can u check this

    SOMEONE ELSES SOLUTION, IM CONFUSED WHY THEY ADD THE MASSES ETC..THEY GET DIFF ANSWER

    and supposedly its the right answer but im not sure why..

    fmax = μs*mg <<agreed assuming m is equal to the smaller mass
    am =ω^2Xm <<agreed since this is just a formula...

    ω=sqrt of k/ (m+M) is the angular frequency <<<< now here is the difference i guess they add m+M i think as a result the answer i got and this answer varies. But which one is right and why!!!!!!!!!!!!

    m*am = μ*mg =>

    k
    ---*Xm = μ*g
    m+M

    Xm = 0.22 m


    I got Xm = 0.0353 m as u can see from my previous post.........


    any help???
     
    Last edited: Jan 11, 2008
  7. Jan 11, 2008 #6
  8. Jan 11, 2008 #7

    Kurdt

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    Omega is the angular frequency and is: [itex] \omega = 2\pi f [/itex] If i recall correctly.
     
  9. Jan 11, 2008 #8

    Kurdt

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    Both methods are fine. If you divide the answer your friend gets by [itex]2\pi[/itex] the answer will be the same as yours. The beauty with yours is it needs less fiddling about.
     
  10. Jan 11, 2008 #9
    O but how my distance is different :O
    so different amplitudes will still give u the same force by the spring?
     
  11. Jan 11, 2008 #10


    how does Dividing or Multiplying by [itex]2\pi[/itex] give u the same answer...like with angles i know if u add [itex]2\pi[/itex] it is the same angle when using sin or cos or tan but how come you can multiply the value of an amplitude by [itex]2\pi[/itex]


    like how will an amplitude of 0.03 or 0.2 BOTH result in the smaller mass falling off.................? i thought there would be a unique solution
     
  12. Jan 11, 2008 #11

    Kurdt

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    There is a unique solution. The flaw with the second method you put up is that omega is incorrectly defined. It is missing a factor of [itex] 2\pi[/itex]. When you use the correct definition you obtain the same answer as in the first method. Try method 2 again with:

    [tex] \omega = 2\pi \sqrt{\frac{k}{m}} [/tex]
     
  13. Jan 11, 2008 #12
    are u 100% sure

    cause in my course notes this is waht i have.............


    F= ma = -m(ω^2*Xm) = -(m*ω^2)*Xm = -kx

    k = mω^2

    ω= [tex] \sqrt{\frac{k}{m}} [/tex]


    no [tex] 2\pi [/tex]
     
    Last edited: Jan 11, 2008
  14. Jan 11, 2008 #13

    Kurdt

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    No you are correct I'm talking rubbish. I don't know where I got that from. If you give me a few minutes I'll go over it properly since I was trying to do too many things at the same time.
     
  15. Jan 11, 2008 #14
    ok thank you so muchhhhhh i will be waiting i have been stuck why mine or perhaps his is wrong for awhile now :(
     
    Last edited: Jan 11, 2008
  16. Jan 11, 2008 #15

    Kurdt

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    First of all sorry I've turned this into a total mess. Now what is wrong with the second method is that you have calculated the acceleration for the two masses on the spring then when you calculate the force you assume only the mass of the small block. What it needs to be is the mass of the two. So:

    [tex] F = (M+m)A\omega^2 = \mu m g [/tex]

    and that will be consistent with the first method.
     
  17. Jan 11, 2008 #16
    ok i think i shoudl have said this before the second method is the prof's solution...and he said its from the sol'n manual but so its wrongg:S???


    by the way thanks alot for taking time to look at this question for me!!!!!
    ------------------------------------

    F= kx

    F = friction force

    therefore

    7.064 = (200)Xm

    Xm = 0.0353 m
    -------------------------------

    is right???
     
    Last edited: Jan 11, 2008
  18. Jan 11, 2008 #17

    Kurdt

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    I hate to have to do this and feel so awful since its entirely my fault (I've been having a bad day). The book is correct and your method must be wrong but I can't see how yet. The derivation you used in post 5 is fine.

    I've been utterly useless in a most embarrassing way and for that I apologise.
     
  19. Jan 12, 2008 #18
    hmm ok
     
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