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Applications of SHM difficult

  1. Apr 14, 2015 #1
    1. The problem statement, all variables and given/known data
    question.png

    2. Relevant equations
    the equation is given in the question

    3. The attempt at a solution
    i tried replacing the 2/3 pi rad as the cos argument and i obtained 2. but the mark scheme says that the position is 3cm above AB.

    in the next part, we have a phase diff of 4/3 pi and i again obtained 2 but the MS still says 3cm above AB. can someone help. am i doing a mistake here.
     
  2. jcsd
  3. Apr 14, 2015 #2

    BvU

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    Hello PS,

    Could you type out the essential, concise, complete problem statement and select which of the equations you want to use in solvig it / And then post your work, please ?

    My magnifying glass doesn't allow me to read the small print, and besides, it would take away too much time away from helping others.
     
  4. Apr 14, 2015 #3
    could yiou try dragging the picture into another tab and enlarge it.
    i uploaded it with an appropriate size but it's not appearing correctly here.

    it's question 3 part b at
    http://www.docstoc.com/docs/173453412/9702_w10_qp_43 [Broken]
     
    Last edited by a moderator: May 7, 2017
  5. Apr 14, 2015 #4

    BvU

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    And what is the problem statement when you reproduce it ? (see the guidelines: it helps you too -- to focus and order your thinking)
     
  6. Apr 14, 2015 #5
    i could work out the in-between questions.
    for the part i'm having problems, i replaced the angle given in the equation given in the question and i obtained 2. the answer says it's 3cb above AB.

    i checked again and there don't seem to be a problem with my solution, in my opinion
     
  7. Apr 14, 2015 #6

    BvU

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    See the guidelines. O&O.
     
  8. Apr 14, 2015 #7
    i did it and still can't understand the problem
     
  9. Apr 14, 2015 #8

    andrevdh

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    Make a rough drawing of the displacement - time graph of X and post it please.
     
  10. Apr 14, 2015 #9
    i already understand how the graph is. its a -cos with height 4
    question1.png
     
  11. Apr 14, 2015 #10

    andrevdh

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    How would the graph look with respect to the previous one if it was leading with say π radians?
    That is the phase difference between the two is π radians, with the 2nd one leading?
    Maybe draw in pencil and post the photo.
    This might be clearer to you: P4140015.JPG
     
    Last edited: Apr 14, 2015
  12. Apr 14, 2015 #11
    question1.png
    in red.

    also, i don't know how to do it by calculation. i tried but did not get the asnwer as i said previously. try to help with this too
     
  13. Apr 14, 2015 #12

    andrevdh

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    Look at the photo in my previous post.
    Do that make any sense to you?
     
  14. Apr 14, 2015 #13
    could you relate it to the type of graph i drew please
     
  15. Apr 14, 2015 #14

    andrevdh

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    The radius line is rotating at a constant speed anti-clockwise.
    It projection on the y-axis is the displacement of the piston.
    At time zero X is at the bottom.
     
  16. Apr 14, 2015 #15
    yeah, but how does that make the position of Y be 3cm above AB.
    and the same is obtained if the angle is 4pi / 3

    could do it it by calculation please. this part actually scores only 1 mark, so i think it should be very simple but i can't get it.
     
  17. Apr 14, 2015 #16

    andrevdh

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    The phase difference between the two is Φ = (2π)/3 rad.
    That is the meaning of the other two lines.
     
  18. Apr 14, 2015 #17
    but how do i obtained the distance travelled / position of the piston?
    AB and CD are separated by 8cm. the amplitude of the oscillation is 4.0cm.

    when phase difference is pi, when one is at AB, the other is at CD, right?
     
  19. Apr 14, 2015 #18

    andrevdh

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    That would be if the phase difference was π, which it is not.
     
  20. Apr 14, 2015 #19
    yeah, but its 120, it's greater than 90, so, it should be greater than 4cm and it's less than 180, so, the distance should be less than 8cm


    could you tell me how to do this by calculation please. i really need to understand this quick
     
  21. Apr 14, 2015 #20

    andrevdh

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    Yes, so the radius line is 120 degrees before that of X, that is Y is at the two o'clock position in my drawing.
     
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