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Finding the period of a vibrating string

  1. Dec 3, 2011 #1
    Hello all,

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

    A vibrating string 50.0 cm long is under a tension of 1.00 N. The results from five successive stroboscopic pictures are shown. The strobe rate is set at 5000 flashes per minute, and observations reveal that the maximum displacement occurred at flashes 1 and 5 with no other maxima in between.

    (a) Find the period, frequency, and wavelength for the traveling waves on this string.

    EDIT:
    (b) How fast is point P moving when the string is in position 3?



    2. Relevant equations

    * A wave pattern travels with constant speed a distance of one wavelength λ in a time interval of one period T.
    f=1/T
    λ=v/f
    v=ωk
    ω=2[itex]\pi[/itex]/T
    k=2[itex]\pi[/itex]/λ

    3. The attempt at a solution

    My main concern is regarding the period -- I understand how to compute the other values thereafter. I am a little unclear as to why certain values are used in computing the period.

    For instance, we are given that over 60 seconds, 5000 flashes occur, therefore 60/5000. Then, in the solution, that value is multiplied by 4. I'm having a hard time conceptualizing why the number 4 is used as oppose to 5, especially since the picture depicts 5 different moments in time.

    From that point, once I have that value, I would assume it to consequently be the period, as that is the amount of time for one wavelength to occur. This is incorrect, however, though I don't understand why.

    Really just having a hard time conceptualizing some little things before I can progress with the question. Any help would be greatly appreciated.

    EDIT:
    For part b, I set v=kω. Using k=2[itex]\pi[/itex]/λ I solved for k, and I solved for ω using ω=2[itex]\pi[/itex]/T and then multiplied the two values together to solve for v, however I'm getting an outrageously high velocity, and I don't believe it is correct. Also, my method doesn't take into account the displacement from the amplitude of 1.5cm, though I'm not sure how to take it into account in this instance.

    Thank you!
     

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    Last edited: Dec 3, 2011
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  3. Dec 3, 2011 #2

    Delphi51

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    The change in time from a peak to a trough is 5 - 1; this is half a period.

    (b) is not asking for the wave speed; rather for the perpendicular speed of the medium. I suggest writing the A*sin(ωt) function for the displacement; then differentiate it to get the function for the velocity. My compliments on your instinct that the amplitude has to have an important effect on the speed!
     
  4. Dec 3, 2011 #3
    Ok, so I'm assuming the A*sin(ωt) comes from A*sin(kx)*sin(ωt), but I'm still unclear about why the sin(kx) portion of the equation is not taken into account, or what happens to it for that matter. But that being said, if I differentiate that, I get -A*cos(ωt). Do I simply set this equal to v? I'm also not sure what my value for A is, given that the string is at its equilibrium, so my thought would be that A = 0?

    I still haven't quite grasped these wave equations...
     
  5. Dec 3, 2011 #4

    Delphi51

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    I don't understand the sin(kx); clearly the waveform in the picture just has one sine.
    Like this: http://en.wikipedia.org/wiki/Amplitude#Formal_representation

    When you differentiate with the chain rule you will get a factor of ω on your
    V = A*cos(ωt).

    You can estimate the value of amplitude A on the picture. Definitely not zero!

    At position 3, the rope is at zero displacement [sin(ωt) = 0] so cos(ωt) will be at its maximum.
     
  6. Dec 3, 2011 #5

    Delphi51

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    I just realized that in the Wikipedia link they forgot the ω! Also, k and b are zero because your wave has no horizontal or vertical shift from the sine wave. I taught high school math and physics for many years and I can vouch for the basic formula of a wave as a function of time is displacement y = A*sin(ωt + k) + b

    Okay, you need the kx if you are interested in the displacement as a function of distance along the path of travel. I suppose in this problem you technically should use the whole formula and have a particular value of x substituted. That would make the derivative of that factor zero. And anyway, the speed will be at its maximum when the amplitude is zero at any value of x.

    Maybe you can help me understand how they get y = A*sin(kx)*sin(ωt) from
    y = A*sin(ωt - kx). According to my trig book sin(A-B) = sin(A)cos(B) - cos(A)sin(B) so your book's A*sin(kx)*sin(ωt) is only the first of TWO terms.
     
    Last edited: Dec 3, 2011
  7. Dec 3, 2011 #6
    I'm afraid I don't quite see it. Using sin(A-B)=sin(A)cos(B)-cos(A)sin(B) I get sin(ωt)*cos(kx) - cos(ωt)sin(kx)?

    What confuses me is if I use the equation given in my book, sin(kx)*sin(ωt) and since k=0, and sin(0)=0, then the rest of my equation becomes 0, whereas in the form you provided, if k=0, it is inconsequential to the rest of the problem.

    If I do as you suggested and use the equation from the book and assign a numerical value for x, the derivative of that portion becomes 0 as you said, but then for the entire derivative, I get cos(ωt)sin(kx)?

    Hm, I think it's definitely the equation itself that's confusing me. I don't think I understand its constituents.
     
  8. Dec 3, 2011 #7

    Delphi51

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    It just clicked with me that we have a standing wave, which can be thought of as the sum of two waves going in opposite directions. The Wikipedia article on standing waves
    http://en.wikipedia.org/wiki/Standing_wave#Mathematical_description
    derives y = A*cos(ωt)*sin(kx) for this situation. At point P, we are 1/4 of the way to a full wavelength along the x axis, so kx = 90 degrees. That gives us
    y = A*cos(ωt)
    That makes sense only if we take the "flash 1" as being t = 0 where we have maximum displacement y = 1.5 cm = A.
    So our formula is y = 1.5*cos(ωt) for the time variation at point P.
    Differentiate that to get the velocity, dy/dt
    "flash 3" is clearly when the wave is 1/4 of the way through its period, so ωt = 90 degrees.

    Your book's y = A*sin(ωt)*sin(kx) is very similar, applying to a rope wave that is flat when ωt = 0. Just a matter of mapping flashes 1 to 5 given in the question so "flash 3" is t = 0, flash 1 is negative 2 flash time intervals, and so on. Then you could use your formula in the same way as the Wikipedia one.

    So sorry I confused both of us on this one! I just wasn't thinking of it being a standing wave. You must remember that the formula applies only to standing waves and has no constants to adjust timing so you have to adjust your clock to make it fit.
     
    Last edited: Dec 3, 2011
  9. Dec 3, 2011 #8

    PeterO

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    Just an explanation of the sentence in red.

    When we count sheep, we start at 1, and the 5th one is number 5. Same with any other set of objects.

    When we count time, we start at zero, so the 5th instant is at the end of the 4th time interval.

    Imagine watching the second hand on an analogue clock [the old fashioned ones] as it sweeps from exactly 2:30 until 5 seconds later.
    At the start the second hand points straight up. [1]
    One second later it points at the first mark around the face [2]
    One second later the second mark [3]
    one second later the third mark [4]
    one second later the 4the mark [5]
    then finally one second later it points at the 5th mark [the 1 o'clock position] [6]

    For the 5 second interval, the hand aligns with 6 markings on the clock face.
     
  10. Dec 3, 2011 #9

    PeterO

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    Next red.

    The period is the time taken for the oscillation to move from any posotion, back to that position.

    While these 5 images were taken, the "high point" on the left hand side [point P] has moved to the bottom. It has to continue oscillating until it is back at the top for a full period.

    Similarly, the low point on the right has reached the top, but we have to wait for it to get back to the bottom for a full period
     
  11. Dec 3, 2011 #10
    I suppose the reason why I'm confused about this is because looking at the first node at x=0 to the third and last node, it looks to me as if one 'cycle' has been completed and that particular point has returned to its original position.
     
  12. Dec 3, 2011 #11
    I appreciate all of the help thus far! I'd be completely lost without it.

    Regarding point P being 1/4 of the way through its period, I guess I'm still confused about the period. If it is already 1/4 of the way along, then by dividing the image into fourths, shouldn't 4/4 be at the third and final node, suggesting what is depicted as one entire wavelength, from node 1 - node 3?
     
  13. Dec 3, 2011 #12

    Delphi51

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    P is 1/4 of the way through the wavelength, not the period. P is at a distance, not a time. Yes, one entire wavelength from the left side of the diagram to the right side. Three nodes including the two ends. Nothing to do with the period, though.

    To find the period, we have to look at the time, indicated by the colored ropes for flashes 1 through 5. In these 4 flash intervals, black to red, the pattern has gone half way to a full period, which would have the rope back to the starting position (black rope, labelled flash 1). It will take 4 more flashes (from the red flash 5) to get back to the initial position. Total of 8 flash intervals in the period.
     
  14. Dec 3, 2011 #13

    PeterO

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    Point P is is moving up and down, not left to right.

    When the first flash is taken [t=0] it is at maximim height. When flash number 2 goes, P has moved down to the point on the green wave at the top of the peak - just to the right of the 2.

    When the 3rd flash goes, P has moved down to a point on the axis, below its original position.

    When the 4th flash goes off, P has moved down to the turquoise wave, just near the 4

    When the 5th flash goes off, P has moved to the lowest point on the pink wave, close to the 5.

    This 5th flash is half way through an oscillation - half a Period after the start.

    If the "flim" had caught the next 4 flashes, you would not really "see" them since image number 6 would be exactly on top of image number 4; Image number 7 would be on top of image #3; image number 8 would be on top of image #2 and image number 9 would be on top of image #1.
    Image #9 would be 1 full period after image #1. Image #5 is one half period after image #1, and image #3 is one quarter period after image #1
     
    Last edited: Dec 3, 2011
  15. Dec 3, 2011 #14

    PeterO

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    Make sure you are reading my corrected post #13 - I got the colours out of order.
     
  16. Dec 4, 2011 #15
    Ok. So, for P to have traveled one entire wavelength, four more flashes -- after what we are shown -- must occur so that it is back at it's original position depicted. If that's the case, why can we say it is 1/4 of its way through a wavelength, and not 1/8? I'm just not seeing it. So, I don't understand why we can say it is at 90 degrees canceling out our sin(kx).

    And then from there, if we have 1.5*cos(ωt), is the correct derivative -1.5*ω*sin(ωt) = v? And now, solving for v, what values do I use for ω and t? ω=2[itex]\pi[/itex]/T? And is my t simply 60/5000?
     
    Last edited: Dec 4, 2011
  17. Dec 4, 2011 #16

    PeterO

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    I don't like your use " for P to have traveled one entire wavelength".

    even if this were a very long slinky with a travelling wave moving through it, no point on the spring travels along a wave length.

    have a look at this applet:

    http://www.walter-fendt.de/ph14e/stwaverefl.htm

    Click on start.

    an incident wave [in red] arrives from the left. once it reaches the fixed end of the string a reflected wave [in blue] returns. The black dots on the string show the resultant movement each particle on the string will have.

    Note that the points only move up and down, they don't move along the wave.

    Once the blue wave has extended all the way along the string, click on, and thus remove, the incident wave and reflected wave boxes on the right.

    What you will then see is a string like in your problem, but actually showing 3 times the length.

    Note that the points are only moving up and down.

    One period is how long [in time!] it takes for a particle to move from a given position/orientation back to that position/orientation.
    eg: highest point to highest point or lowest point to lowest point or mean position, moving up to mean position, moving up etc.

    You should recognise that the 5 pictures in your problem, taken over 4 time intervals, represent the first half of a cycle - that is why we would need 4 more time intervals for a full cycle - a full period.
    One quarter of the way through a cycle, and the string is actually in its original position - however most of it will be moving at high speed at the time.
    You might like to play with the pause/resume button to see if you can "stop" the applet in the "flash #3" position.

    Let me know if you need further explanation after you have seen this.

    Sorry about the delay in answering: time zones and all that - I am in Australia
     
  18. Dec 4, 2011 #17
    No need to apologize, thank you so much for the help! It is immensely appreciated.

    I understand all of what you wrote in your last message up until "One quarter of the way through a cycle, and the string is actually in its original position..." After 4/4 cycles, won't point P just be at the bottom red line?
     
  19. Dec 4, 2011 #18

    PeterO

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    Those references were to the applet.

    Before you start running it, the string is straight and horizontal.

    Coming back to your original problem/posting. if we name the lines shown Black, Green, axis, Blue, Red, then while these pictures were taken, point P will move vertically.
    It starts on Black - moves to green, then axis, then blue then Red. That does not represent a full cycle; a full cycle will have it back where it starts.
    A full cycle is:
    Black → green → axis → blue → red → blue → axis → green → black

    As you can see, Red is half way through - half a cycle - and the axis [the first time] is 1/4 cycle through.

    5000 flashes per minute, and 8 flashes being one cycle, means the period is 8/5000 of a minute [which needs to be converted to seconds]

    With regards to "the wavelength of the traveling wave on the string" in the original post:

    While we see a standing wave on the string, there is an underlying traveling wave causing it - this is well illustrated in that applet I referred you to.
     
  20. Dec 4, 2011 #19
    I think I get it! So, 1/4 of the way through the cycle corresponds to 90 degrees, and that's why sin(ωt)=1 at that point, leaving A*ω for the answer to the velocity at that point. My question at this point, is that since the derivative of the position equation gives us -A*ω*sin(ωt) for the velocity, why is the negative sign disregarded in the answer? Seems to be a problem I've run into several times now.

    Also, what should I be thinking of in terms of what the (ωt) actually represents in the equation. I think I get confused a lot with these problems because I think I need to calculate separate values for ω and t, but that doesn't seem to be the case.
     
  21. Dec 4, 2011 #20

    PeterO

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    The minus sign will probably have to do with direction.

    If the original position is 1 [ie +1] and it is 1 unit up., then 1/4 period later, when it is travelling down, the velocity will be negative, meaning down.

    The original question asked "how fast". Fast is a speed reference, not a velocity reference, so any sign would be ignored anyway.

    I have never explicitly used the "ωt" concept - I work it out using geometry and circular motion.

    If you re-run that applet and reduce it to a standing wave only, you will hopefully be able to picture it as a multiple looped skipping rope, rotating. I calculate the speed as it travels straight down at T/4 as simply the speed of the loop traveling in the circle. Otherwise I would be taking the derivative of sin(2πt/T), which is probably sin(ωt) anyway, but I have never written it with the ω symbol.

    BTW: if you get a long, reasonably massive rope [2cm diameter, 6m long] you can try having people skip in it while you rotate it at "double speed" forming two loops like in your original post [though the rope rotates rather than springs back and forth]. It is great to watch that two people skipping in that rope actually jump at different times. Helps implant the idea of this string in your mind.
    With "normal" skipping, we rotate the rope quite slowly so that a single, large loop is formed. Spin the rope faster and it drops into this double loop formation.
    Spin even faster and you will get a triple loop, but the sound of the rope whistling through the air means you are unlikely to get any volunteers willing to get even near it, let alone try to skip in the rope.
     
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