# Difficult Wave Problem

1. Aug 23, 2007

### PFStudent

1. The problem statement, all variables and given/known data

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,
$$y(x, t) = {y}_{m}sin\left(kx - \omega t + \phi\right)$$
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.)

2. Relevant equations

Wave Equation – Transverse Displacement

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

3. The attempt at a solution

$$x = 0{{.}}m$$

$$\phi = ?$$

I first recognize that $x = 0{{.}}m$, reducing my transverse displacement equation to,

$$y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)$$

From here I let the transverse velocity,

$$u = {v}_{y}$$

Just to be easier to recognize.

From looking at the graph I note that at: $t = 0{{.}}s$ the transverse velocity, ${v}_{y} = -4{{.}}m/s$.

Then,

$$y(x, t) = {y}_{m}sin\left(- \omega t + \phi\right)$$

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

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

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

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

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

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

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

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

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,

$${v}_{y_{max}} = 5{{.}}m/s$$

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. Aug 23, 2007

### 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.

3. Aug 24, 2007

### 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.

Last edited: Aug 24, 2007
4. Aug 24, 2007

### PFStudent

Hey,

Thanks for the response.

I do not quite follow how,

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

Drops to,

$$v_{y_{max}}(x, t) = \omega y_m$$

I recognize that the argument,

$$- \omega t + \phi\right = \theta$$

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, $v_{y}(x, t) = 5{{.}}m/s$?

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

$$\theta = 0{{.}}Degrees$$

or

$$\theta = 180{{.}}Degrees$$

or

$$\theta = 360{{.}}Degrees$$

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

$$v_{y_{max}}(x, t) = \omega y_m$$

Thanks, any help is appreciated,

-PFStudent

5. Aug 25, 2007

### PFStudent

Hey,

Am I correct in saying that for,

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

The expression inside the cosine can be let equal theta,

$$\theta = - \omega t + \phi\right$$

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

$${v}_{y_{max}} = 5{{.}}m/s$$

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

$$\theta = 0{{.}}Degrees$$

or

$$\theta = 180{{.}}Degrees$$

or

$$\theta = 360{{.}}Degrees$$

Thanks,

-PFStudent

Last edited: Aug 25, 2007
6. Aug 25, 2007

### 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 $$\omega>0$$ and $$y_{m}>0, \omega{y}_{m} = 5$$

Last edited: Aug 25, 2007
7. Aug 25, 2007

### 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 $$\phi$$

8. Aug 25, 2007

### PFStudent

Hey,

In order for,

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

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

$$v_{y_{max}}(x, t) = +\omega y_m$$

Let,

$$\theta = - \omega t + \phi$$

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,

$${v}_{y_{max}}(x, t) = +\omega{y}_{m}$$

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,

$$v_{y_{max}}(x, t) = +\omega y_m$$

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

$${v}_{y}(x, t) = 5{{.}}m/s$$

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
9. Aug 25, 2007

### 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.

Last edited: Aug 25, 2007
10. Aug 25, 2007

### 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 $$\phi$$