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What exactly does 'x-vt' mean in the wave equation?

  1. Oct 7, 2015 #1
    So the equation is
    f(x,t) = f(x-vt,0) = A sine k(x-vt).

    Now Since x is displacement of wave from origin, t is time taken for that,wouldn't 'x-vt' always be equal to 0 ?
    If x-vt is 0,
    Y = A sine k(x-vt) will also be zero.
    So am I writing the equation wrong?
     
  2. jcsd
  3. Oct 7, 2015 #2

    davenn

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    Hi there Jayxl
    welcome to PF :smile:

    no, as you said x is the displacement = velocity "v" times t (time) so it can only be 0 if there is no wave propagation

    the - sign represents a forward ( rightward ) wave propagation of the wave (x-vt)
    if a + sign then its a leftward ( backwards) propagating wave (x+vt)


    Dave
     
  4. Oct 7, 2015 #3

    DrDu

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    What kind of waves are you considering?
    x isn't the displacement of the wave whatever, but the point in space whence the displacement is measured.
     
  5. Oct 7, 2015 #4

    davenn

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    I could get out of my depth very quickly here :wink:
    not really my field of interest .... am not familiar with the 0 in the f(x-vt,0), am wondering if that defines the
    displacement at the origin time which really means f(x-vt,0) = f(x) ?

    hope some one more knowledgeable chimes in :)


    Dave
     
  6. Oct 7, 2015 #5

    davenn

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    which is the same as the displacement from origin @ time t, when velocity =v

    you cant measure the displacement of a point without referring to where it has been displaced from
     
  7. Oct 7, 2015 #6

    DrDu

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    That's why I was asking about which waves we are talking. For example, I could consider transversal waves on a string. Then if the extension of the string is in x direction and the tranversal displacement is in y direction then f=y (if the undisplaced string is at y=0).
     
  8. Oct 7, 2015 #7

    davenn

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    ahh ok

    I may be wrong ( as said, not my forte) I was understanding that this was for longitudinal waves travelling through/along a medium
    ie, along your guitar string, not the traverse motion if the string .... can it be used to describe both ?


    D
     
  9. Oct 7, 2015 #8

    DrDu

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    Of course it can. But in general, x refers to the undisplaced position ##x=x_\mathrm{undisplaced}## and f would be ##f=x_\mathrm{displaced}-x_\mathrm{undisplaced}##.
     
  10. Oct 7, 2015 #9

    davenn

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    OK so how does that fit in with

    To displace any function f(x) to the right, just change its
    argument from x to x-x0, where x0 is a positive number.

    If we let x0 = v t, where v is positive and t is time, then the displacement
    increases with increasing time

    am always willing to learn :smile:

    D
     
  11. Oct 7, 2015 #10
    x is not the displacement of the wave from the origin. x is just the x coordinate of the point where we are measuring the displacement of the waving medium from its normal value. This could be a displacement in the same direction of the x-coordinate (longitudal waves), or in a different direction (transverse waves), or even a pressure (sound waves).

    The solution of the wave equation f(x,t) = Asin (k (x-vt)) gives you the values of the displacement for all values of x and t, so saying x-vt has to be 0 makes no sense
     
  12. Oct 7, 2015 #11
    For a given position [itex]x[/itex] and time [itex]t[/itex], you evaluate the function [itex]f(x,t)[/itex] at those values to determine the height of the wave.

    This is different from [itex]v[/itex], the speed of propagation of the wave. Consider yourself moving to the right with displacement defined by [itex]x(t)=v' t +x_0[/itex]. Then, plugging this into [itex]f(x,t)[/itex] you can see that [itex]f(x(t),t)[/itex] is constant in time if and only if [itex]v' = v[/itex]. Thus we conclude that the speed of the wave is indeed [itex]v[/itex].
     
  13. Oct 7, 2015 #12

    atyy

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    f(x,t) means f is a function of x and t, eg. x2t3.

    However, that is too general for a solution to the wave equation, which is why you wrote f(x,t) = f(x-vt,0), which means that f could be any function of x-vt, eg. (x-vt)3 + exp(x-vt).

    If you have a function f(x), then f(x+a) is the shape shifted to the left by a.

    So f(x-vt,0) mean that as t increases, vt increases, and the shape shifts to the right by vt. In other words the shape travels to the right as time increases.
     
  14. Oct 7, 2015 #13

    jtbell

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    I've never seen the general function written that way, either. I've always seen it as just f(x-vt). Perhaps the OP could give us a link to where he saw it, or maybe someone else recognizes it.
     
  15. Oct 9, 2015 #14

    morrobay

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    With y = A sin (kx - ωt) the calculator should be on radians for sin , correct ?
    I just made up some numerical values for : A =1. λ =3m. k = 2.09 rad/unit distance , frequency = 4 cy/sec , v = 12m/s . ω = 25.12 rad/s
    At t0
    x0 = 10m
    At t1 = 4 sec
    x1 = 58 m
    So y = Asin(kx-ωt) = y = sin (121.2-100.4)
    With sin in radians = .955
    With sin in degrees = .355
     
    Last edited: Oct 9, 2015
  16. Oct 10, 2015 #15

    jtbell

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    y=f(x-vt) simply means that x and t appear only in the combination x-vt, possibly with other constants "surrounding" that combination (loosely speaking). Here's a simple example: y=(x-2t)2. To see that 2 is really a velocity, set t to a couple of different values, and make graphs of y versus x for each of them.

    Setting t=0 gives y=x2. The graph is a parabola with its vertex at the origin (x=0).

    Setting t=1 gives y=(x-2)2. The graph is a parabola with the same shape as before, but with its vertex on the x-axis at x=2.

    Setting t=2 gives y=(x-4)2. The graph is a parabola with the same shape as before, but with its vertex on the x-axis at x=4.

    Clearly y=(x-2t)2 represents a parabola moving in the +x direction with speed v=2.

    What about ##y=A \sin (kx-\omega t)##, a common description of a sinusoidal wave? Rewrite it a bit: $$y=A \sin \left[ k \left( x-\frac{\omega}{k}t \right) \right] \\
    y = A \sin [k(x-vt)]$$ where ##v=\omega/k##. So this is also in the form y=f(x-vt).

    Now put in some numbers. Let A=1, ##k=\pi/5## and ##\omega=0.4\pi##, so $$y = \sin \left( \frac{\pi}{5} x - 0.4 \pi t \right)$$ As we did for the parabola above, make three graphs of y versus x, setting t=0, t=1 and t=2 respectively. You should get three sinusoidal graphs, with wavelength ##\lambda = 2\pi/k = 10##, shifted from one to the next in the +x direction by v = ω/k = 2 units.
     
    Last edited: Oct 10, 2015
  17. Oct 10, 2015 #16

    robphy

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    Amplifying jtbell's comments...

    Check out https://www.desmos.com/calculator/bjt6dleg5h and the screenshot below.

    Let u=x-vt, with v>0.
    When t=0, then u=x (for all x).
    Consider the snapshot of a disturbance in space at t=0 (the dotted graph).
    At a later time t=1, that disturbance has advanced to the right by a displacement vt.
    (I tried to show this on a faux 3-d plot. Imagine a stack of xy-planes for each time t, with the disturbance function drawn on it...each plane drawn slightly displaced to minimize overlap.)

    Pick any value of U (say U=0)... and consider the value of the function at f(0), call it A.
    So, A is the disturbance at u=0 (that is when x-vt=0 or when x=vt).
    At time t=0, A is the disturbance at x=0 (since u=0).
    At time t=1, A is the disturbance at x=v*1 (since u=0).
    At time t=2, A is the disturbance at x=v*2 (since u=0).
    That is to say, the "disturbance at x=0 when t=0" is advancing to the right with velocity v.

    Now pick another value of U (say U=0.1)....
    and consider the value of the function at f(0.1), call it B.
    So, B is the disturbance at u=0.1 (that is when x-vt=0.1 or when x=0.1+vt).
    At time t=0, B is the disturbance at x=0.1 (since u=0.1).
    At time t=1, B is the disturbance at x=0.1+v*1 (since u=0.1).
    At time t=2, B is the disturbance at x=0.1+v*2 (since u=0.1).
    That is to say, the "disturbance at x=0.1 when t=0" is advancing to the right with velocity v.

    And so on for each value of U that you choose (that is, for each location x at time t=0).

    Thus, with v>0,
    "f(x-vt)" describes the disturbance pattern in space (i.e. at locations x) moving to the right (as t increases, x must increase to keep x-vt=constant).
    and
    "f(x+vt)" describes the disturbance pattern in space (i.e. at locations x) moving to the left (as t increases, x must decrease to keep x+vt=constant).

    https://www.desmos.com/calculator/bjt6dleg5h
    bjt6dleg5h.png
     
    Last edited: Oct 10, 2015
  18. Oct 11, 2015 #17

    DrDu

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    Quite interesting discussion. A pitty that the OP didn't show up again!
     
  19. Oct 11, 2015 #18

    morrobay

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    Last edited: Oct 12, 2015
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