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Difficult Wave Problem

  1. Aug 23, 2007 #1
    1. The problem statement, all variables and given/known data

    4. Figure 16-31 shows the transverse velocity [itex]u[/itex] versus time [itex]t[/itex] of the point on a string at [itex]x = 0[/itex], as a wave passes through it. The wave has form,
    [tex]
    y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)
    [/tex]
    What is [itex]\phi[/itex]?
    (Caution: A calculator does not always give the proper inverse trig function, so check your answer by substituting it and an assumed value of [itex]\omega[/itex] into [itex]y(x, t)[/itex] and then plotting the function.)

    [​IMG]

    2. Relevant equations

    Wave Equation – Transverse Displacement

    [tex]
    y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)
    [/tex]

    3. The attempt at a solution

    [tex]
    x = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m
    [/tex]

    [tex]
    \phi = ?
    [/tex]

    I first recognize that [itex]x = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m[/itex], reducing my transverse displacement equation to,

    [tex]
    y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)
    [/tex]

    From here I let the transverse velocity,

    [tex]
    u = {v}_{y}
    [/tex]

    Just to be easier to recognize.

    From looking at the graph I note that at: [itex]t = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}s[/itex] the transverse velocity, [itex]{v}_{y} = -4{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s[/itex].

    Then,

    [tex]
    y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)
    [/tex]

    [tex]
    \frac{\partial}{\partial t}\left[y(x, t)\right] = \frac{\partial}{\partial t}\left[{y}_{m}sin\left(- \omega t + \phi\right)\right]
    [/tex]

    [tex]
    {y}_{t}'(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
    [/tex]

    [tex]
    {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
    [/tex]

    I recognize that even with substituting, [itex]t = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}s[/itex] and [itex]{v}_{y} = -4{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s[/itex]; there will still be two unknowns, [itex]{y}_{m}[/itex] and [itex]\phi[/itex].

    So I figured the transverse acceleration would be useful to solve for [itex]{y}_{m}[/itex],

    [tex]
    \frac{\partial}{\partial t}\left[{v}_{y}(x, t)\right] = \frac{\partial}{\partial t}\left[-\omega{y}_{m}cos\left(- \omega t + \phi\right)\right]
    [/tex]

    [tex]
    {v}_{y_{t}}'(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi \right)
    [/tex]

    [tex]
    {a}_{y}(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi\right)
    [/tex]

    However, the transverse acceleration attempt seems to have left me no where, as I still have no way of solving for [itex]{y}_{m}[/itex] or [itex]\phi[/itex].

    I recognize that there is significance in noting from the graph that the maximum value for,

    [tex]
    {v}_{y_{max}} = 5{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s
    [/tex]

    However, I am not quite sure exactly how that helps…

    Yea, so I’m pretty stuck. Any help is appreciated.

    Thanks,

    -PFStudent
     
    Last edited: Aug 24, 2007
  2. jcsd
  3. Aug 23, 2007 #2

    learningphysics

    User Avatar
    Homework Helper

    I haven't checked all your work, but it seems like from your equation for [tex]v_y[/tex], [tex]v_{y_{max}} = \omega * y_m[/tex]

    So solve for [tex]y_m[/tex] using 5m/s.
     
  4. Aug 24, 2007 #3
    On figure 16-31 you can see two extrema.
    These define one half period of the wave.
    The position of the Zero-crossing in-between the extrema give you a direct evaluation of the phase.
    No need for long calculations, you simply need a ruler.

    Note, however, that using the vertical scale of the graphics could provide you a more precise position of the extrema.
    By this method, then, inverse trig functions migh be necessary since you will use the y-coordinates to spot t-coordinates.

    I don't understand the comment about calculators, since you don't even need inverse trig function, only simple arithmetics.
     
    Last edited: Aug 24, 2007
  5. Aug 24, 2007 #4
    Hey,

    Thanks for the response.

    I do not quite follow how,

    [tex]
    {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
    [/tex]

    Drops to,

    [tex]
    v_{y_{max}}(x, t) = \omega y_m
    [/tex]

    I recognize that the argument,

    [tex]
    - \omega t + \phi\right = \theta
    [/tex]

    However, where is [itex]\theta[/itex] measured?

    On this graph ([itex]v_{y}[/itex](x,t) vs. t) or on (y(x,t) vs. t)?

    If it is measured on this graph ([itex]v_{y}[/itex](x,t) vs. t), then is it measured at the "flat point" on the curve where, [itex]v_{y}(x, t) = 5{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s[/itex]?

    That is to say, because at that point on the graph the curve becomes flat, is the theta measurement,

    [tex]
    \theta = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    or

    [tex]
    \theta = 180{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    or

    [tex]
    \theta = 360{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    That is where I am pretty much stuck in understanding your equation,

    [tex]
    v_{y_{max}}(x, t) = \omega y_m
    [/tex]

    Thanks, any help is appreciated,

    -PFStudent
     
  6. Aug 25, 2007 #5
    Hey,

    Am I correct in saying that for,

    [tex]
    {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
    [/tex]

    The expression inside the cosine can be let equal theta,

    [tex]
    \theta = - \omega t + \phi\right
    [/tex]

    So for the point in the graph where the curve hits a maximum at,

    [tex]
    {v}_{y_{max}} = 5{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s
    [/tex]

    Would the angle for that maximum point on the graph be,

    [tex]
    \theta = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    or

    [tex]
    \theta = 180{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    or

    [tex]
    \theta = 360{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees
    [/tex]

    Thanks,

    -PFStudent
     
    Last edited: Aug 25, 2007
  7. Aug 25, 2007 #6

    learningphysics

    User Avatar
    Homework Helper

    the function y = A cos(wx), or any sinusoidal function (that isn't vertically shifted)... has a maximum of |A|... because -1<=cos(angle)<=1

    And yes, that maximum(or minimum) occurs when the angle is 0,180,360.... at 180n (where n is any integer).

    so assuming [tex]\omega>0 [/tex] and [tex]y_{m}>0, \omega{y}_{m} = 5[/tex]
     
    Last edited: Aug 25, 2007
  8. Aug 25, 2007 #7

    learningphysics

    User Avatar
    Homework Helper

    Since you know that [tex]v_{y} = -5 cos(-\omega{t} + \phi)[/tex] and you know that [tex]v_{y}=-4[/tex] at t=0, you can calculate [tex]\phi[/tex]
     
  9. Aug 25, 2007 #8
    Hey,

    Thanks for the reply.

    In order for,

    [tex]
    {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)
    [/tex]

    To drop to [itex]{v}_{y_{max}}[/itex],

    [tex]
    v_{y_{max}}(x, t) = +\omega y_m
    [/tex]

    Let,

    [tex]
    \theta = - \omega t + \phi
    [/tex]

    Where [itex]cos\theta = -1[/itex],

    Cosine theta must equal negative one to cancel with the negative in front of omega in [itex]{v}_{y}(x, t)[/itex], in order to maximize [itex]{v}_{y}(x, t)[/itex].

    Thus yielding,

    [tex]
    {v}_{y_{max}}(x, t) = +\omega{y}_{m}
    [/tex]

    So, then [itex]v_{y_{max}}[/itex] only occurs when theta is 180n (Where n is any odd integer)

    I refer to odd integers because at even integers, [itex]cos\theta = +1[/itex], which would instead yield the minimum, [itex]v_{y_{min}} = -\omega{y}_{m}[/itex]

    Ok....so,

    I agree that the maximum or minimum tranverse velocity occurs at 180n (where n is any integer).

    In addition, the maximum transverse velocity only occurs at 180n (where n is any odd integer) as this would require, [itex]cos\theta = -1[/itex] which would minimize the cosine function and then cancel with the negative one and therefore maximize the, [itex]v_{y}(x, t)[/itex] function. Yielding,

    [tex]
    v_{y_{max}}(x, t) = +\omega y_m
    [/tex]

    Ok then, by looking at the maximum on the graph where,

    [tex]
    {v}_{y}(x, t) = 5{\textcolor[rgb]{1.00,1.00,1.00}{.}}m/s
    [/tex]

    As you have mentioned, the angle is 180 degrees.

    Thanks,

    -PFStudent

    [EDIT 1] I edited this post to reflect your two subsequent posts, thanks for the help.
     
    Last edited: Aug 25, 2007
  10. Aug 25, 2007 #9

    learningphysics

    User Avatar
    Homework Helper

    In this case, since [tex]v_{y} = -5cos(-\omega{t} + \phi)[/tex] with a -5, intead of positive 5... the maximum occurs at 180, 540, 900, .... ie 180 + 360n.

    for the velocity to take on the maximum, the cosine has to be -1... then -5(-1) = 5, the maximum.
     
    Last edited: Aug 25, 2007
  11. Aug 25, 2007 #10

    learningphysics

    User Avatar
    Homework Helper

    The maximum occurs here when the angle within the cosine is 180 degrees.

    Plug in t=0 and [tex]v_{y} = -4[/tex] to solve for [tex]\phi[/tex]
     
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