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Velocity Derivative of a Sinusoidal Wave (Counter-Intuitive)

  1. May 21, 2015 #1
    What's the matter:
    So, I think I have some skills when it comes to differentiation after taking calculus 2 last semester, but when it starts to intertwine with physics, and interpreting physical phenomenon through equations, It appears I could use some help. Anyway, the problem that I got hung up on is admittedly simple by anyone's standards; introductory classical physics, more specifically dealing with waves.

    If you take the dy/dt of position function, in this case a sinusoidal wave function, you find it's velocity function.
    y(x,t)=xmsin(xk(0)-ωt) you should get:
    U=-xmωcos(-ωt)
    However, that's not what happens.
    BTW. Two different sources attest to this being the right answer.
    http://[URL=http://s172.photobucket.com/user/Alpha_Scope/media/Derivative%20Question_zpsryocyscm.jpg.html][PLAIN]http://i172.photobucket.com/albums/w15/Alpha_Scope/Derivative%20Question_zpsryocyscm.jpg [Broken]
    Derivative%20Question_zpsryocyscm.jpg

    It's counter-intuitive to me since this seems like a simple derivative, and I'm unable justify it through the logic that one uses when manipulating mathematics to fit physical interpretation. As I mentioned previously, I'm still getting the hang of all this. Perhaps a little sooner with your help.

    Anyone care to shed some light?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 21, 2015 #2

    Orodruin

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    Your image does not show, you cannot link to things which are in your yahoo account and expect them to be visible to others.

    Furthermore, your derivative is missing the kx of the cosine argument.
     
  4. May 21, 2015 #3
    Okay, should I expect it to appear now? I did so intentionally because x is said to equal 0.
     
  5. May 21, 2015 #4

    Orodruin

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    This is the correct derivative if x = 0 (or a multiple of pi/k).
     
  6. May 21, 2015 #5
    Apparently this is a typo, cant have ##\pi/t## there but the correct is ##\pi/5##. I am not sure though what you dont understand. Why dy/dt is the velocity of the point of matter at x=0? It is by the definition of velocity, we know that the point of matter at x=0 does an oscillation given by ##y_0=f(0,t)=-4sin(-\pi t/5)##, hence the velocity of this oscillation is ##\frac{dy_0}{dt}=\frac{df(0,t)}{dt}=-4(-\frac{\pi}{5})cos(-\frac{\pi t}{5})##.
     
  7. May 21, 2015 #6
    Okay, maybe there ain't a problem here after all. I found this solution (if I should even call it one) to this problem reiterated a couple different times, so I thought I made a mistake. I guess nothing's the matter, since this ain't my bad.

    THANKS GUYS!
     
  8. May 21, 2015 #7
    Ok i see now the term ##\pi/t## if it was correct it would mean that the velocity goes to zero as time grows larger (goes to infinity) which we know it isnt correct for an oscillating point of matter. I wonder how can it be that many different sources reproduce this obvious(both mathematically and physically) mistake.
     
    Last edited: May 21, 2015
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