What is the phase of the resultant wave when x = t = 0?

[wrongusername]

Homework Statement

Two harmonic waves are described by (pic attached). What is the phase of the resultant wave when x = t = 0?

A. 3
B. 1
C. 4
D. 2

Homework Equations

$$y=Asin(kx-\omega(t)+\phi)$$ (Single wave equation)
$$y=2Asin(kx)cos(\omega(t))$$ (Standing wave equation)

The Attempt at a Solution

I originally chose D because the phase angle in the second equation was -2. However, then I thought about that fact that the question asked for the resultant wave. Thus, I was thinking that since both x and t are 0, the first wave equation reduces to 0 and the second to 7m sin(-2). Thus, the resultant wave equation should be only 7m sin(-2). Where do I got from there? (Is that even the correct thing to do?)

 

[rl.bhat]

Find the resultant y1 and y2. While doing it you have to use some trigonometric simplification.
sinC +sinD = 2*sin[(C+D)/2]*cos[(C-D)/2]
Then decide the phase difference.

 

[wrongusername]

Find the resultant y1 and y2. While doing it you have to use some trigonometric simplification.
sinC +sinD = 2*sin[(C+D)/2]*cos[(C-D)/2]
Then decide the phase difference.

What would be the phase difference after I've done that? (I haven't solved for the resultant wave yet, I'm just wondering how I get the phase difference after I do so)

 

[rl.bhat]

Just solve it. The phase difference is 1.

 

[wrongusername]

Just solve it. The phase difference is 1.

Thank you very much , but what I meant was, how I get the phase difference from the resultant wave function (in other words, in $$y=Asin(kx-\omega(t)+\phi)$$ I know that the phase difference is $$\phi$$, but I wasn't quite so sure what the phase is in the resultant wave equation).

I think the resultant wave equation is $$(14m)sin(5\frac{x}{m}-100\frac{t}{s}-1)cos(1)$$. I see two one's in that equation, so I still am not completely sure how to find the phase difference. Also, the question asks for the phase when x and t are both 0. How do time and position come into play? I thought the phase was a constant?

More help would be much appreciated

 

[rl.bhat]

When you write
y = A*sin(kx- ωt + φ), φ is the starting point of the oscillation. And that is called phase. In that sense the phase of the resultant wave is 1 rad. It is not the phase difference. Even in the problem they have asked phase.
We consider phase difference when we compare two different waves. In that case, in the given problem, one wave has zero phase and other has 2 rad. The phase difference between them is 2 rad.

 

[wrongusername]

When you write
y = A*sin(kx- ωt + φ), φ is the starting point of the oscillation. And that is called phase. In that sense the phase of the resultant wave is 1 rad. It is not the phase difference. Even in the problem they have asked phase.
We consider phase difference when we compare two different waves. In that case, in the given problem, one wave has zero phase and other has 2 rad. The phase difference between them is 2 rad.

Just solve it. The phase difference is 1.

I'm all confused now

So what is, and how do I find, the phase of the resultant equation I'm trying to find?

 

[rl.bhat]

In my post#4, I should have mentioned it as phase, not the phase difference.

 

[wrongusername]

In my post#4, I should have mentioned it as phase, not the phase difference.

Thank you... can you clarify how to calculate or to find that for the phase?

 

[rl.bhat]

Thank you... can you clarify how to calculate or to find that for the phase?

I don't know what you want to calculate.

 

[wrongusername]

I don't know what you want to calculate.

The phase.

 

[rl.bhat]

Your resultant wave is
y = 14*sin(5x/m - 100t/s -1)
When x = 0, t = 0. phase is -1 rad, and displacement is y = -14*sin(1 rad) = -11.78 m.

 

[wrongusername]

Your resultant wave is
y = 14*sin(5x/m - 100t/s -1)
When x = 0, t = 0. phase is -1 rad, and displacement is y = -14*sin(1 rad) = -11.78 m.

Why is there no cosine part to the resultant wave equation? Is phase -1 or +1 rad?

 

[rl.bhat]

Why is there no cosine part to the resultant wave equation? Is phase -1 or +1 rad?

Yes. You are right. When x = 0 and t = 0, the amplitude of the resultant wave is 14*sin(1 rad)*cos(1 rad)
And phase is -1 rad.