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I Doubts on wavefunction conditions

  1. Jul 3, 2016 #1
    I'm facing some difficulties in using "boundary conditions" in a simple wavefunction.

    The wavefunction I'm considering is $$\xi(x,t)=A sin (k x \pm \omega t +\psi)$$
    The minus or plus are for progressive or regressive waves. The indipendent parameters are 4: ##A##, ##k##, ##\omega##, ##\psi##.

    The first doubt I have is: can ##A## be negative or is it defined as positive in any case? (In the following examples I willl consider a progressive wave, so the wavefunction is ##\xi(x,t)=A sin (xk - \omega t +\psi)## ).
    • Consider a oscillating rope: all I know is that one end of the rope moves according to the function ##f(t)=B sin(\gamma t)##, with ##B>0## and ##\gamma>0##. What I would do here is impose that ##\xi(0,t)=f(t)##. Since there is a minus in front of ##\omega##, before imposing ##\xi(0,t)=f(t)## I think I must rewrite ##\xi(0,t)=A sin (-\omega t+\psi)=-A sin(\omega t-\psi)=f(t)##, from which I can conclude ##A=-B## and ##\gamma=\omega## (and then also ##\psi=0##). So ##A## would be negative.
    • Another example of ambiguity on the sign of ##A## is when I'm given ##k## and ##\omega## and the two conditions ##\xi(\bar{x},\bar{t})=s_0## and ##\frac{\partial \xi}{\partial t} |_{(\bar{x},\bar{t})}=v_0## . That means ##A sin (k \bar{x} -\omega \bar{t} +\psi)=s_0## and ##A \omega cos(k \bar{x} -\omega \bar{t}+\psi)=v_0## ##\implies## ##A^2=s_0^2+(\frac{v_0}{\omega})^2##. So I get ##A^2## but how to choose the sign of ##A## in this case?
    The other doubt is : in previous examples I supposed that the wave is progressive. In some exercises I see that it is not even specified. Is the choose of sign plus or minus infront of ##\omega## a fifth indipendent condition? That is, can this be determined if I know all the four parameters or not? My guess would be no, since two wave can have identical parameters but still be progressive or regressive.

    I really appreciate any suggestion regarding these two doubts!
     
  2. jcsd
  3. Jul 3, 2016 #2

    Simon Bridge

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    Re: the sign of A:
    How does the amplitude of the wave depend on the value of A?
    What is the physical definition of "amplitude" in the case of real wave described by that equation? (ie. think about the physical meaning of the parameters.)

    Bullet points: what values can ##\psi## have?
    What reason do you have to assert the sign of ##\omega##?
    A=-B means that A<0 iff B>0. Is it the case that B>0 ... be careful in your descriptions, not everything you can write mathematically is physically meaningful. Try describing the physical situation in words before writing out the algebra.

    Hint: ##A\sin(\omega t) = -A\sin(\omega t + \pi)##
     
    Last edited: Jul 3, 2016
  4. Jul 12, 2016 #3
    Thanks a lot for the reply! I thought about the problem again but still have some unclear points.

    The physical definition of ##A## is the distance between the poisition of equilibrium and maximum excursion of the rope. That means that ##A##, as amplitude, is necessarily positive. Is this consideration valid?

    On the other hand I'm sure that ##\omega## is positive by itself too.

    ------------------------------------------------------------------------------------------------------------------------------------------------------
    Concerning the first example I made, I tried in a different manner avoiding to use ##-A=B##, but getting the same result

    The condition is ##\xi(0,t)=A sin(-\omega t+\psi)=B sin (\gamma t)##. Since I cannot conclude that ##-\omega=\gamma## (because one of them would be negative) I can rewrite this as $$-A sin(\omega t +\psi)=B sin(\gamma t) \implies A sin (\omega t +\psi +\pi)= B sin( \gamma t)$$ $$\implies A=B \, , \, \omega=\gamma \, , \, \psi=-\pi$$ So $$\xi(x,t)= B sin(kx - \gamma t-\pi)=-B sin(kx-\gamma t)$$

    In this way the number in front of the since function in my wavefunction is a negative number, so in the end it is like the "amplitude is negative". But is this way to proceed correct or is there a conceptual mistake?

    ------------------------------------------------------------------------------------------------------------------------------------------------------

    If ##A## must be positive then also my second example would be solved because in that case I have to take the solution with positive sign. Would that be correct?
     
  5. Jul 13, 2016 #4

    Simon Bridge

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    By definition, the amplitude must be positive. A negative amplitude has no physical meaning. But the displacement from equilibrium at some position may be positive or negative and it will have some relationship to the amplitude.
     
  6. Jul 16, 2016 #5

    Simon Bridge

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    Note..
    Completely generally, the equation is of form ##y(x,t) = a\sin (bx+ct+d)## where all constants may be any real number.
    But we need to be able to extract physical meaning from the result.
    So we may define physical quantites as your example: ##A,k,\omega## as the amplitude, wave number, etc of the physical wave.
    Then it follows that ##A=|a|##
    If we insist on the standard form for the equation to be ##y=A\sin (kx-\omega t + \delta)## then we have to select consistent values for A, k etc that give the same wave as the a,b,c values from before.
    But we are lazy and we just jump right to the standard form, ignoring, at first, the requirement that some quantities cannot take negative values.
    This sloppiness leads to ambiguity mathematically, but its ok physics because we have Nature to provide context.
    For instance... if we get a negative amplitude, this tells us that the phase offset ##\delta## is out by pi.
     
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