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Mechanical waves

  1. Jan 8, 2008 #1
    [SOLVED] Mechanical waves

    1. The problem statement, all variables and given/known data
    Hi all.

    Please see the last page in this .ppt (problem 15.66):

    http://web.utk.edu/~kamyshko/P232/Problems_13_15.pdf

    3. The attempt at a solution

    Here are my answers - I would be grateful, if you would read them through and comment/help where necessary:

    a) Since there are 5000 flashes/min, there's 83,3 flashes/s. I want to find the time it takes for 5 flashes, which is 5/83,3 = 0,06 s. This is the half of the period T (half the unit-circle?), so the period T = 2*0,06 s = 0,12 s.

    The frequency is T^(-1), and to find the wavelength, I will use that the string is fixed at both end and since the string is one wavelength long, we can use that the wavelength lambda = L.

    b) Since there is only one point not moving, it is the second harmonic (first overtone).

    c) To find the speed of the travelling waves, I use that v = lambda*f.

    d) To find the speed of point P at position 1, I differentiate y(x,t) w.r.t. t and insert x = L/4 (since P is at the top of the first crest so L/4 horizontal distance from starting point) and t = 0.

    To find the speed of point P at position 3, I do the same as above.

    e) To find the mass of the string, I use that v = sqrt(F/mju), where mju is mass/unit length.

    - Thanks in advance,

    sincerely Niles.
     
    Last edited: Jan 8, 2008
  2. jcsd
  3. Jan 8, 2008 #2

    Kurdt

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    You seem to have completed this problem with no trouble. The only point i'd raise is for part d. In position 1 its at its maximum amplitude so the speed is minimum. At position 3 it will be maximum.
     
  4. Jan 8, 2008 #3
    Ahh yeah, of course. So in position 1 the velocity is 0 and in position 3 the velocity is omega (angular frequency) * amplitude. So there's no need to start differentiating at all?
     
  5. Jan 8, 2008 #4

    Kurdt

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    Of course that comes from differentiating and setting the position to zero. I mentioned it because I think the purpose of the question was to show that you know the speed is maximum when the amplitude is 0 and minimum when the amplitude is a maximum.
     
  6. Jan 9, 2008 #5
    In c), when they are asking for the transversal velocity - can I use v = lambda*frequency? I mean, does this also apply to finding the transversal velocity of a standing wave?

    And if I wish to find the velocity of the point P in the first position (question d) using v(x,t), which values for x and t should I plug in?
     
  7. Jan 9, 2008 #6

    Kurdt

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    Yes you can. You need it for part e.

    I assume you have some sort of equation of the form:

    [tex] y(x,t) = A_n\sin\left(\frac{2\pi x}{\lambda_n}\right)\cos(\omega_nt) [/tex]

    The x position will be 1/4 the wavelength and the time will be 0.
     
  8. Jan 9, 2008 #7
    I use this equation for a standing wave:

    y(x,t) = 2*A*sin(kx)*sin(wt).

    When I diff. wrt. time, I get v(x,t) - I find that the velocity at position 1 is w*A and velocity at position 3 is negative? This doesn't see correct.

    For position 1: v(0.125 ; 0) = 2*A*sin(k*0.125)*cos(0), where k = 2*pi*f.
     
    Last edited: Jan 9, 2008
  9. Jan 9, 2008 #8

    Kurdt

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    The equation you have assumes different starting conditions. Thats the problem with the differential equation is you always have to factor in the starting conditions or assumed starting conditions. You can of course just phase correct it.
     
  10. Jan 9, 2008 #9
    Ok, now it's correct for position 1, but for position 3 it gives me that v_y = -0,95*A*w. I can't see why it isn't just -A*w? I use the time t = 0.036 (the time it takes for three pictures).

    To phase i correctly, I added a -pi/2.
     
    Last edited: Jan 9, 2008
  11. Jan 9, 2008 #10

    Kurdt

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    What are you using for omega? remember that [itex] \omega = 2\pi f[/itex].

    Just noticed as well in hind sight that the time from the first to the last photo will be 4 times the time between frames. This is because if frame 1 is t = 0 then frame 5 will be 4 times the time between frames. Therefore the time from 1 - 3 will be 0.024s.
     
  12. Jan 9, 2008 #11
    Of course, I see. Now it's correct - thanks :-)
     
  13. Jan 9, 2008 #12

    Kurdt

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    Sorry I didn't spot it earlier.
     
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