Difficult Wave Problem

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  • #1
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Homework Statement



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.)

http://img251.imageshack.us/img251/1050/chpt16p4fig1631jpgaz1.jpg [Broken]

Homework Equations



Wave Equation – Transverse Displacement

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

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 by a moderator:

Answers and Replies

  • #2
learningphysics
Homework Helper
4,099
6
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.
 
  • #3
1,256
2
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:
  • #4
170
0
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.

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
 
  • #5
170
0
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:
  • #6
learningphysics
Homework Helper
4,099
6
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:
  • #7
learningphysics
Homework Helper
4,099
6
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]
 
  • #8
170
0
the function y = A cos(wx), or any sinusoidal function... 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]

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:
  • #9
learningphysics
Homework Helper
4,099
6
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:
  • #10
learningphysics
Homework Helper
4,099
6
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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