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[Physics waves] Spring and Block on inclined plane

  1. Sep 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A block weighing 9.81N oscillates on a plane inclined of 60 degrees attached to a spring whose constant
    point is k=100N/m and the length is to equilibrium l0 = 0.6 m. While the block goes to the
    top losing speed , a timer is started ( to t0 = 0) when the block passes through the
    point where the potential energy of the spring assembly system equals the kinetic energy of the block (U0 =K0) and
    measuring the K(t0)= K0 = 2J. Note that the inclined plane is also the potential energy
    Total block -spring system is given by
    U = 1/2* kx2 .
    The block kinetic energy is
    K =1/2* m v2 .
    The total energy of the system is preserved and satisfies the relationship :
    E = U + K =1/2 *kA2.
    a) What are the initial position xo and vo initial speed of the block ?
    b) What is the maximum length of the spring when it oscillates ?



    2. Relevant equations


    3. The attempt at a solution
    a)
    since K0 = U0
    and K0 = 2J then U0 also equals 2J.

    Then I can use those 2 equations:

    W = 9.81N so m = 1kg

    U = 1/2* k*x2 .
    2J = 1/2* 100*x2 .
    x= 0.2m
    K =1/2* m v2
    2J = 1/2 * 1kg * v2
    v = 2m/s

    b)
    we know that l0 = 0.6
    W = sqrt(k/m)
    W = sqrt(100/1) = 10rad/s

    E = U + K =1/2 *kA2.
    E = 2J + 2J =1/2 *kA2.
    2J + 2J =1/2 *100*A2.
    A = 0.282843

    I am pretty much stuck here and not sure how to proceed to find the maximum length of the spring when it oscillates.
     
  2. jcsd
  3. Sep 16, 2015 #2

    andrevdh

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    Why are you not including the gravitational potential energy of the block?
     
  4. Sep 16, 2015 #3
    It is added to find A.
    U is the potential energy
     
  5. Sep 16, 2015 #4
    Anyone can help me with this?
     
  6. Sep 17, 2015 #5

    andrevdh

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    Since the block is oscillating on an incline one would think that the gravitational potential energy would change as is goes up and down the incline. I thought U is the potential energy stored in the spring?
     
  7. Sep 17, 2015 #6
    Thats the part I dont understand.
    How can I calculate that ?
     
  8. Sep 17, 2015 #7

    andrevdh

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    I should think that you need to choose a zero level and then calculate the vertical height above or below this point with the given angle.
     
  9. Sep 17, 2015 #8

    andrevdh

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    If you choose the zero level for the gravitational potential energy, UG, at the to point the the total energy of the system would be .... joule at every other point in the oscillation.
     
  10. Sep 17, 2015 #9
    1/2* kx2 .+1/2* m v2 .=1/2 *kA2.

    I know A,k,m
    and I dont know x and v, how am I supposed to do this to get max length of spring?
     
  11. Sep 18, 2015 #10

    andrevdh

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    lo seems to be the length of the spring at equilibrium. The max length would then be when it is at the bottom of the oscillation. So it would just be lo + A. That is the block oscillates with amplitude A about the equilibrium position. Here I am assuming the block hangs from the spring rather than sitting on top of it. Problem is I don't think you calculated A correctly. It should really help you if you make a drawing.
     
  12. Sep 18, 2015 #11
    How else would I calculate A?
    and yes that is how it is, block hangs from the spring.
     
  13. Sep 19, 2015 #12
    I still need help :(
     
  14. Sep 20, 2015 #13
    I still need help :(
     
  15. Sep 20, 2015 #14
    Can anyone help me ?
     
  16. Sep 21, 2015 #15

    andrevdh

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    You can calculate the phase constant associated with the SHM if the zero gravitational potential is chosen at the point where the energy stored in the spring is equal to the kinetic energy of the block. Which would then enable you calculate its initial position.
     
    Last edited: Sep 21, 2015
  17. Sep 21, 2015 #16

    andrevdh

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    You could also determine the phase at which the velocity is zero and use it to determine the displacement at that stage, which would then be the amplitude.
     
  18. Sep 22, 2015 #17
    v0 = −Aω sin(ωt0 + ϕ)
    0 = −Aω sin(ω*0 + ϕ)
    0 = −Aω sin( ϕ)
    ω = sqrt( 100/ 1) = 10
    0 = −A*10 sin( ϕ)
    I don't know A and ϕ, how could I possibly find 2 unknown variable ?
     
  19. Sep 22, 2015 #18

    andrevdh

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    You have assumed incorrectly that the velocity is zero at t = 0 or to. I am losing the line of reasoning since this is stretching over such a long period of time. What do you mean by this?
    should it be "Note that the bottom of the inclined plane is also the zero potential energy level"?
     
    Last edited: Sep 22, 2015
  20. Sep 22, 2015 #19
    Should of said this: " The total potential energy block -spring system is given by U = 1/2* k*x2 . "

    But then how do I know which "t" to use ?
    What im wondering is why the A I calculated in the beginning is wrong ?
     
  21. Sep 23, 2015 #20

    andrevdh

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    Well, I was thinking along the line that in the energy conservation game it has a certain amount of kinetic energy, 2J, at this point and it will travel upwards until it is converted into potential energy, but not only that stored in the spring, also that stored in the block-earth configuration or gravitational potential energy. As to the time I thought if you could somehow determine just the phase at which the velocity is zero, then you could use the same phase to determine the position of the oscillator, which would be its amplitude. The problem seems to boil down to how do gravity influence an shm oscillator. It does not change its frequency, but how do it factor into its amplitude?
     
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