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Graphing a vertex plane help

  1. Apr 3, 2005 #1
    1.In the figure, describe the motion of particles X and Y at the instant.
    What will be the shape of the string after 1/4 cycle?
    I don't know which direction that the crests are travelling as there are two sources.
    2.The tetrahedron BCDH is cut off from the cube and is then placed on top of the solid ABDEFGH as shown in the figure. The face BCD of the tetrahedron coincides with the face BAD of the solid ABDEFGH such that vertex H of the tetrahedron moves to the position of V and vertex C coincides with A. The faces BHD and BVD of the new solid lie on the same plane.

    What's the projection of point F on the plane BDH?
     

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  3. Apr 3, 2005 #2

    quasar987

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    What do you mean 2 sources? There's a vibrator to the left and a wall to the right. So the waves are coming from the left.

    But it doesn't matter in which direction they are traveling. The answer to this question is based on the fact that in such a vibration (called a normal mode), each point on the robe describes a VERTICAL simple harmonic motion of SAME FREQUENCY. This means that each point on the rope crosses equilibrium position at the same time!
     
    Last edited by a moderator: Apr 3, 2005
  4. Apr 3, 2005 #3
    I haven't learnt standing wave yet. Would you mind explaining it further so that I can understand it thoroughly? Why won't the wave reflect from the right to the left?
    Do you have any ideas of the second question?
     
  5. Apr 3, 2005 #4

    quasar987

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    Hey, that was funny Halls of Ivy!

    primarygun:

    I didn't read second question. The figure gave me an instant headache :yuck:..

    They do! But the frenquency of oscillation of the vibrator, along with the principle of superposition make it so it appears that they are not.
     
  6. Apr 4, 2005 #5
    Oh, so the wave reflected is out of phrase to the original wave?
    But,in reality, when someone helps me hold one end and I move the another end as if the picture shows, why a straight line does not result in?
     
  7. Apr 4, 2005 #6

    quasar987

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    Yup!

    I'm not sure what you mean.
     
  8. Apr 4, 2005 #7
    Such, a man is holding point P, not letting it move any where.
    Then, another man is flicking the other end up and down, what would you see?
    Moreover, in the figure, are X and Y at their maximum speed at the instant?
     
  9. Apr 4, 2005 #8

    quasar987

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    Depends on the frequency at which you're flicking.

    No. When it the harmonic oscillator speed maximum? Think of a pendulum... or of a block on a spring.

    It is probably to be suposed that in the picture, X and Y are at their point of least velocity ==> maximum amplitude.
     
  10. Apr 5, 2005 #9
    But Y is not on the crest
     
  11. Apr 5, 2005 #10

    quasar987

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    So?

    -----------------
     
  12. Apr 5, 2005 #11

    AKG

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    2. By Pythagoras, you can find the length FH. Consider the triangle VFH. With some trig, you can find the angle at H, <VHF. You want to find the point on the plane BDH such that the line connecting that point to F is perpendicular to the plane. By symmetry, you know this point will occur on VH (i.e. it won't be closer to B than to D, or vice versa). Let this point be P. Well you already know <PHF = <VHF, and the angle <FPH = 90 (right angle, perpendicular). Thus it's easy to find the angle <PFH. You have a right triangle, knowing all the angles and one side length (FH). You can figure out the other side lengths with this information, and this should be enough to let you place P on the plane. Certainly, once you find PH, knowing that P is on the line HV, you can say where P is in terms of it's distance from H on the line HV. There might be other ways to express the position of the projection of F on BDH, and it may take a little more trig. to find those expressions (nothing hard, nothing requiring finding more values, just setting up equations and plugging in), but the given question doesn't specify the terms in which P needs to be expressed, so this should suffice.
     
  13. Apr 6, 2005 #12
    So projection of F to that plane is on a point of VH?
    Construct a line perpendicular from F to VB cut at Z( I named it) and from Z on VB to draw a perpendicular on the plane VBHD. The line on the plane VBHD naming is ZY
    i.e. the two lines are drawn to find the angle between the two planes.

    Then, The projection of F is a point on ZY, isn't it?
    And I found that ZY is not on the plane VBHD but is the extension part of the plane.
     
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