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The maximum and minimum transverse speeds of a point at an antinode

  1. Feb 28, 2008 #1

    TFM

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    [SOLVED] The maximum and minimum transverse speeds of a point at an antinode

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

    Adjacent antinodes of a standing wave on a string are 15.0cm apart. A particle at an antinode oscillates in simple harmonic motion with amplitude 0.850cm and period 0.0750s. The string lies along the +x-axis and is fixed at x = 0.

    the speed of the two travelling waves are 4.00ms

    Find the maximum and minimum transverse speeds of a point at an antinode.

    2. Relevant equations

    I'm not sure

    3. The attempt at a solution

    Any help, I would have thought that they would have the same speed as the wave, ie 4 ms

    TFM
     
  2. jcsd
  3. Feb 28, 2008 #2
    Simple harmonic motion means that the particle's vertical position can be given by the function y(t)=Asin(wt) where A is the max amplitude and w is the angular speed (2 pi / period). The deriative of that function will give you the speed as a function of time. Find the max and min of speed of that, and you are done. (Another hint: max of a absolute value of a cosine is 1 and min is 0).
     
  4. Feb 28, 2008 #3

    TFM

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    [tex] v = A\omega cos (\omega t) [/tex]

    I get

    [tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(\omega t) \omega[/tex]

    I'm not sure what you meant by the cos part (I understand max absolute value = 1, minimuim = 0)

    TFM
     
  5. Feb 28, 2008 #4
    That seems about right. The cosine just means that the velocity will not be constant over time and will follow a cosine curve. However the question only asks for the maximum and minimum speeds, so all you have to do is find max and min of the function you found.

    Btw, I think you have an extra w at the end.
     
  6. Feb 28, 2008 #5

    TFM

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    Yeah, that extra w shouldn't be there.

    I feel silly, but...

    how do I work out the maximum/minimum of the function - don't you have to differentiate it again and find when it is equal to 0?

    TFM (Rather embarrased :blushing:)
     
  7. Feb 28, 2008 #6
    That's where the maximum and minimum of the cosine comes in. Since the velocity only changes as a function of time. And as time changes the only thing that changes is the cosine. So the function will be at a maximum when cosine as at a maximum and at minimum when consine is at minimum.
     
  8. Feb 28, 2008 #7

    TFM

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    Lets see if I understand, the maximum will be when:

    [tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(1) [/tex]

    And a minimum when:

    [tex] v = 0.0045 * (\frac{2 \pi}{0.075}) cos(0) [/tex]

    ?

    TFM
     
  9. Feb 28, 2008 #8
    Not quite. The velocity will be largest when [cos(wt)] is at it's largest (the whole cosine function, not whats inside the cosine function, since what's inside the function can get very large if time gets very large. But no mater how big (wt) gets, the cosine function never gets larger than a certain value, and it is the value of the whole cosine function that multiplies the other terms in the equation to give the velocity), and the velocity will be at a minimum when cos(wt) is at it's smallest (absolute value). What is the largest value that a cosine function ever reaches? What is the smallest (absolute) value that cosine ever reaces?
     
  10. Feb 28, 2008 #9

    TFM

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    Ah, Should be:

    [tex] cos (\omega t) = 0 [/tex] for minima

    [tex] cos (\omega t) = 1 [/tex] for maxima

    that gives 0 and pi/2, though? do you use 2 pi?

    TFM
     
    Last edited: Feb 28, 2008
  11. Feb 28, 2008 #10
    right, wt=0 -> velocity is max, wt=pi/2 -> velocity is minimum. What are those maximum and minimum velocities?
     
  12. Feb 28, 2008 #11

    TFM

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    I used pi/2 and 2pi as the values, which gave me 2.37 and 0.59, but they were stated as being wrong:cry:

    TFM
     
  13. Feb 28, 2008 #12
    And your calculator is in radian mode right. :)
     
  14. Feb 28, 2008 #13
    You should be able to use zero, just as well as 2pi...
     
  15. Feb 28, 2008 #14

    TFM

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    I'm using excel, which operates inj radians

    I did

    cos(wt) = 0 for minima, giving pi/2
    cos(wt) = 1 for maxima giving 0, 2pi

    then

    using:

    v = Awcos(wt)

    A = 0.00425 m (0.425cm)
    w = 2pi/0.075 (period = 0.075 s)

    [tex] v_m_i_n = 0.00425*(2pi/0.075)*(pi/2) [/tex] = 2.23
    [tex] v_m_a_x = 0.00425*(2pi/0.075)*(2pi) [/tex] = 0.559

    Any ideas?

    TFM
     
  16. Feb 28, 2008 #15
    The pi/2 and 0,2pi are the values of wt that make the cosine function 0 and 1, respectively. By doing what you did, you are saying v = Aw*(wt), as opposed to v=Awcos(wt).
     
  17. Feb 28, 2008 #16

    TFM

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    I did v = Awcos(pi/2), Awcos(0), which gave me a maximum of 0.356, and a minimum of 2.18 x 10^-17m, which is wrong!?!?

    TFM
     
  18. Feb 28, 2008 #17
    Well, I think one of them is right...I'm pretty darn sure the min is zero (which is what 2.18x10^-17 is).

    What are the correct answers?
     
    Last edited: Feb 28, 2008
  19. Feb 28, 2008 #18
    Oh, and your aplitude should be .0085, not .00425, so you would be getting half the correct answer for the max...
     
  20. Feb 28, 2008 #19

    TFM

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    I don't know, masteringphysics only tells you when you're wrong.:rolleyes:

    any ideas?

    TFM
     
  21. Feb 28, 2008 #20

    TFM

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    That makes the answers right:

    0.712, 4.36 x 10^-17

    Thanks,

    TFM
     
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