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Time between sound wave displacements

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    tI2yrCp.png

    2. Relevant equations


    3. The attempt at a solution
    So I set up:

    2.0 nm = (6.0 nm)cos(kx + (3000 rad/s)t1 + phi)
    -2.0 nm = (6.0 nm)cos(kx + (3000 rad/s)t2 + phi)

    My professor said if there's a phi you can usually ignore it so I'm just going to remove them:

    2.0 nm = (6.0 nm) * cos(kx + (3000 rad/s)t1)
    -2.0 nm = (6.0 nm) * cos(kx + (3000 rad/s)t2)

    Simplify them:

    1/3 nm = cos(kx + (3000 rad/s)t1)
    -1/3 nm = cos(kx + (3000 rad/s)t2)

    I need to solve for each time, but we have 3 unknowns and only two equations so I'm not sure what else to do from here.
     
  2. jcsd
  3. Feb 26, 2013 #2

    tms

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    One thing you can do is set [itex]t_0[/itex] to zero. Then you can set [itex]x[/itex] to zero, too.
     
  4. Feb 26, 2013 #3
    But then we get

    s = (6.0 nm) * cos(k*0 + (3000 rad/s)*0)
    s = 6.0 nm * 1
    s = 6.0 nm

    I'm not sure how this helps
     
  5. Feb 26, 2013 #4

    tms

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    You left out the phase constant.
     
  6. Feb 26, 2013 #5
    Oh...I'm not sure when to include the phase constant or not because sometimes it seems negligible and when we are doing problems the professor just leaves it out. But doesn't that add another variable anyway?

    s = (6.0 nm) * cos(phi)
     
  7. Feb 26, 2013 #6

    tms

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    Well, if you want to omit [itex]\phi[/itex], then stick [itex]x[/itex] back in. But then you'll have to worry about [itex]k[/itex], too.
     
  8. Feb 27, 2013 #7
    I don't understand this relationship. Why does x and k re-appear if phi disappears?
     
  9. Feb 27, 2013 #8
    If you set your frame of reference so that x = 0 and t1 = 0, then you have to specify the starting phase through phi in order to get the required initial displacement 2.0 nm.
     
  10. Feb 27, 2013 #9

    tms

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    I meant [itex]t_1[/itex].
     
  11. Feb 27, 2013 #10
    how can I calculate the value of phi to accomodate for the different starting point?
     
  12. Feb 27, 2013 #11
    The starting phase must be such that the 6.0 sin (phase) = 2.0. This follows directly form the description of the problem.
     
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