How Do You Calculate Phase Shift in a Wave Equation?

[PFStudent]

Homework Statement

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

http://img251.imageshack.us/img251/1050/chpt16p4fig1631jpgaz1.jpg

Homework Equations

Wave Equation – Transverse Displacement

<br /> y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)<br />

The Attempt at a Solution

<br /> x = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}m<br />

<br /> \phi = ?<br />

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

<br /> y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)<br />

From here I let the transverse velocity,

<br /> u = {v}_{y}<br />

Just to be easier to recognize.

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

Then,

<br /> y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)<br />

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

<br /> {y}_{t}&#039;(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)<br />

<br /> {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)<br />

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

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

<br /> \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]<br />

<br /> {v}_{y_{t}}&#039;(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi \right)<br />

<br /> {a}_{y}(x, t) = -{\omega}^{2}{y}_{m}sin\left(- \omega t + \phi\right) <br />

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

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

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

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

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

Thanks,

-PFStudent

 

[learningphysics]

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

So solve for y_m using 5m/s.

 

[lalbatros]

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.

 

[PFStudent]

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

So solve for y_m using 5m/s.

Hey,

Thanks for the response.

I do not quite follow how,

<br /> {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)<br />

Drops to,

<br /> v_{y_{max}}(x, t) = \omega y_m<br />

I recognize that the argument,

<br /> - \omega t + \phi\right = \theta<br />

However, where is \theta measured?

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

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

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

<br /> \theta = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

or

<br /> \theta = 180{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

or

<br /> \theta = 360{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

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

<br /> v_{y_{max}}(x, t) = \omega y_m<br />

Thanks, any help is appreciated,

-PFStudent

 

[PFStudent]

Hey,

Am I correct in saying that for,

<br /> {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)<br />

The expression inside the cosine can be let equal theta,

<br /> \theta = - \omega t + \phi\right<br />

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

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

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

<br /> \theta = 0{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

or

<br /> \theta = 180{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

or

<br /> \theta = 360{\textcolor[rgb]{1.00,1.00,1.00}{.}}Degrees<br />

Thanks,

-PFStudent

 

[learningphysics]

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 \omega&gt;0 and y_{m}&gt;0, \omega{y}_{m} = 5

 

[learningphysics]

Since you know that v_{y} = -5 cos(-\omega{t} + \phi) and you know that v_{y}=-4 at t=0, you can calculate \phi

 

[PFStudent]

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 \omega&gt;0 and y_{m}&gt;0, \omega{y}_{m} = 5

Hey,

Thanks for the reply.

In order for,

<br /> {v}_{y}(x, t) = -\omega{y}_{m}cos\left(- \omega t + \phi\right)<br />

To drop to {v}_{y_{max}},

<br /> v_{y_{max}}(x, t) = +\omega y_m<br />

Let,

<br /> \theta = - \omega t + \phi<br />

Where cos\theta = -1,

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

Thus yielding,

<br /> {v}_{y_{max}}(x, t) = +\omega{y}_{m}<br />

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

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

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, cos\theta = -1 which would minimize the cosine function and then cancel with the negative one and therefore maximize the, v_{y}(x, t) function. Yielding,

<br /> v_{y_{max}}(x, t) = +\omega y_m<br />

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

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

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.

 

[learningphysics]

In this case, since v_{y} = -5cos(-\omega{t} + \phi) 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.

 

[learningphysics]

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

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