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Simple math problem

  1. Jun 12, 2005 #1
    Simple wave problem

    Hi guys!
    I'm trying to solve a wave problem for homework and I hope that I can extract all data, once I have my wave function in the form: y(x,t)=A*sin(kx-wt) :biggrin: . The simple problem is to convert the given data: y(x,t)=4.2*cos(0.2x)*sin(300t) into the above form :uhh: . It seems that I've forgotten mathematics at all!! Can you please show me how such a conversion works, because when I tried the f-la for sin(a)*cos(b) I obtain two sin()s when I need only one. :confused:
    It is later asked for the wavelength, frequency, speed, etc. But I hope I can solve this with the "proper stated" wave function. :wink:
    Thanks in advance!
     
    Last edited: Jun 12, 2005
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  3. Jun 12, 2005 #2

    Hurkyl

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    Just a guess, but maybe your wave is the superposition of two sinusoids?
     
  4. Jun 12, 2005 #3
    OooooK, but then my problem becomes more difficult :biggrin:. How to extract the wavelength and frequency of such an equation? Any hints? :uhh:
     
  5. Jun 12, 2005 #4

    Hurkyl

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    Think about what they mean, physically, rather than how to read them off of a sinusoid.

    (Though, I suspect you could get the answer by reading them off of the individual sinusoids, but that's merely educated guessing, and that's not good enough!)
     
  6. Jun 12, 2005 #5
    The problem is that I don't know how to deal with such separated expressions - one for sin (containing "t") and one for cos (containing "x")... If it is a superposition, then how to separate the two functions?!
     
  7. Jun 12, 2005 #6

    Hurkyl

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    I thought this meant that you got an expression for your wave as a sum of two sinusoids!



    Anyways, I'm pretty sure that you don't need to do any algebraic manipulations on this problem -- you should be able to read the wavelength and frequency straight from the expression as given... if you remember what those terms mean.
     
  8. Jun 12, 2005 #7
    That's right and exactly this confuses me :rolleyes:. But leaving the confusion aside, I obtain from [tex]y(x,t)=4.2\cos(0.2t)\sin(300t)[/tex] the function [tex]y(x,t)=2.1(\sin(300t + 0.2t) + \sin(300t - 0.2t))[/tex]. So if [tex]\lambda=\frac{2\pi}{k}[/tex] and I leave the sign, I get [tex]\lambda=10\pi[/tex]. For [tex]f=\frac{\omega}{2\pi}=\frac{150}{\pi}[/tex]. Correct? Because I'm not sure how to deal with the two different signs...
    Moreover, I'm asked to find the speeds. Is it possible to write the equation in this form: [tex]y(x,t)=2.1(\sin(0.2(x+\frac{300}{0,2}t)+\sin(0.2(x-\frac{300}{0,2}t))[/tex] and obtain the speed -> 300/0.2 = 1500?
    Can you please check if the logic is correct and provide some hints if i'm going completely the wrong way :redface:
     
  9. Jun 12, 2005 #8

    Hurkyl

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    This is what I meant by "educated guessing", and it sounds like you've figured it out. (Though, I think you're reading the parameters wrong off of your sinusoids)


    I still advise you figure out how to work out the frequency and the wavelength from first principles, rather than from the canned formulae that work only for sine waves. If your professor gave you this problem, I wouldn't put it past him to put similar problems on tests! And those might not be so easily worked out by educated guessing!

    (Not to mention you could use it to promote your work on this from educated guessing to rigorous proof!)
     
  10. Jun 12, 2005 #9
    I suppose first principles means straight from the given equation, without transforming it in any way. :shy:
    However, being new to this field, makes it a little more difficult for me to understand such hints. Probably if you provide a little more specific tips or advices, I will be able to grasp the idea and solve my problem. :wink:
    Thanks for your time :rolleyes:
     
  11. Jun 12, 2005 #10

    Hurkyl

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    I have -- can you quote the definition of wavelength and frequency, or at least give a reasonable approximation?
     
  12. Jun 13, 2005 #11
    OK, If I take in mind that wavelength is the distance between successive points of equal amplitude and phase on a wave, and frequency is the number of cycles or events per unit time, I take the given equation [tex]y(x,t)=4.2\cos(0.2x)\sin(300t)[/tex] and I see that it has a double frequency (due to cos()*sin()... or at least my teacher presented it so) :). This means that the frequency should be half of the value I formerly gave and the wavelength - double. Am I reading the parameters correct off my sinusoids? :)
     
  13. Jun 13, 2005 #12

    OlderDan

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    The sin(300t) term has all the time dependence. For any constant value of x, y varies in time in proportion to that sin(300t) term. You get the frequency directly from that. There is no "double frequency". The spatial variation of the waveform at some moment in time is completely described by the cos((0.2x) term. Look at a value of x for which the cosine term is zero. Can y ever be anything other than zero at that x by changing t? What about the two separate sinusoids you found. Do they behave the same way as their combination? How do those two sinusiods differ from one another?
     
  14. Jun 13, 2005 #13

    Hurkyl

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    If your professor used those exact words, he's wrong. :tongue2: He probably said it's doubly periodic, meaning it has two different periods. (A period is, minor details aside, something you can add to the inputs that doesn't change the outputs)


    Right -- a slightly different way of saying it is that, no matter where you are, if you move a distance of one wavelength, you get the same amplitude!


    Right. IMHO it's conceptually easier to look for the time per cycle, and then invert to get the frequency.


    (You should also note there's a duality here -- shouldn't everything said about the wavelength also apply in a similar, but not quite the same, way to the frequency?)
     
  15. Jun 14, 2005 #14
    Following the advices of you both, I calculated the values for x when the cos is 0: [tex]x=\frac{\pi}{0.4}[/tex] and [tex]x=\frac{5\pi}{0.4}[/tex], so the distance between two successive knots is [tex]10\pi[/tex]. This gives me the wavelength [tex]\lambda[/tex]. :tongue2:
    The period for one full cycle I obtained in the same way, using the sin this time -> [tex]t=\frac{\pi}{150}[/tex], so the frequency is the reciprocal: [tex]f=\frac{150}{\pi}[/tex].
    The speed I find when I multiply them both - [tex]v=f\lambda=1500[/tex].
    Did I get the point now? :biggrin:
     
  16. Jun 14, 2005 #15

    OlderDan

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    Fortunately, you calculated a full period of the cosine function, not the interval between successive zeros. The full spatial period is the wavelength. You should recognize that the function you are working with describes a standing wave, which is the superposition of two traveling waves, one moving in the positive x direction and the other in the negative x direction. I had not noticed the typo error you made earlier.
    should have said (when combined)

    [tex]y(x,t)=4.2\cos(0.2x)\sin(300t) = 2.1\left[\sin(300t + 0.2x) + \sin(300t - 0.2x)\right][/tex]

    One of the last two terms is a wave traveling in the +x direction, and the other is a wave traveling in the -x direction with angular frequency [tex] \omega = 300 [/tex] and wave number [tex] k = 0.2 [/tex]
     
  17. Jun 14, 2005 #16
    Oh, yeah, sorry for the mistake :tongue2: ...
    As far as I've understood you correctly, you say that everything is OK now, right? :bugeye:
     
  18. Jun 14, 2005 #17

    OlderDan

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    It appears to me that you have correctly identified the wavelength and frequency, so the velocity is correct. I just wanted to make sure you understand the connection between that standing wave and the decomposition into the two traveling waves.
     
  19. Jun 14, 2005 #18

    Hurkyl

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    Your work looks perfect.
     
  20. Jun 15, 2005 #19
    Thank you guys, you really helped me! :smile: It seems that I've got the point now...
    However, I have one or two more questions about this problem (I've not stated them yet), but I think I will post them in a new thread, because they are not "simple math problems" any more :rolleyes:
    So for now thank you and see you in the next thread :biggrin:
     
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