# Homework Help: Phase Shift of two sinusoidal waves

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1. Mar 24, 2017

### Techno_Knight

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

Two sinusoidal waves in a string are defined by the wave functions
y1 = 2.00 sin (20.0x – 32.0t)
y2 = 2.00 sin (25.0x – 40.0t)
where x, y1, and y2 are in centimeters and t is in seconds.

(a) What is the phase differencebetween these two waves at the point x = 5.00 cm at t = 2.00 s?

(b) What is the positive x value closest to the origin for which the two phases differ by ±π at t = 2.00 s?
(At that location, the two waves add to zero.)

2. Relevant equations

3. The attempt at a solution

I solved (a) (I'm just typing the abridged version to save some time, I know about the units):

Δφ = (25*x - 40*t) - (20*x - 32*t) = 5*x 8*t
We put in t = 2s and x = 5 cm and that's it.

But (b)'s giving me some trouble, because I'm not sure exactly what I'm meant to use. My official solutions manual, gives this answer:

(b) The sine functions repeat whenever their arguments change by an integer number of cycles, an integer multiple of 2π radians. Then the phase shift equals ±π whenever Δφ = π + 2nπ, for all integer values of n.
Substituting this into the phase equation, we have:

At t = 2.00 s, π + 2nπ = −(5.00 rad/cm)x + (8.00 rad/s)(2.00 s)

The smallest positive value of x is found when n = 2:

Okay, I understand what he says about the phase increasing by 2π with every cycle. And, after he gets his result, I understand why n = 2 ( n = 0 & n = 1 give larger numbers, n = 3 gives us a negative answer, so n = 2 gives us the smallest positive value of x). What I don't get, is how he came to the conclusion of Δφ = π + 2nπ. Also, Δφ = +- π, so where is the (-)?

As I've said before, english isn't my native tongue, so I'm basically missing some "key words", and I'm moving "faster" than I'd like to (i.e. not so much time spent reading the chapters), so any help would be apreciated.

Thanks!

2. Mar 24, 2017

### haruspex

What are the units of Δφ?

3. Mar 25, 2017

### Techno_Knight

I bypassed them to save some time.

4. Mar 25, 2017

### haruspex

Yes, but my point is that from what you wrote you would have ended up with 9 radians, yes?
The -π corresponds to +π with a different value of n. -π + 2nπ = π + 2(n-1)π.

5. Mar 25, 2017

### Techno_Knight

Yup, that's correct. Sorry I didn't add this, it's just that it didn't really have anything to do with the second part, so i just wrote down the "barebones" version.

Oh, okay. That's one down. Now, could you make it a tad simpler/clearer why he takes Δφ = π + 2nπ ? I'm having a bit of trouble understanding the previous paragraph ("The sine functions...we have:"). Dunno if it's because I'm missing something or if it's a "english is not my native tongue" related issue, but I'd appreciate a "dumbed down" version of it. I get the first part (with each cycle the phase increases by 2π), but afterwards I get a bit lost.

6. Mar 25, 2017

### haruspex

Ok, but it would be a bit unusual to give a phase difference as "9 radians". Since a phase difference of φ is indistinguishable from one of φ+2π it would be more usual to quote a value either in the range -π to π or in the range 0 to 2π (or if the unsigned difference is wanted, just 0 to π).
Perhaps my note above explains that too. A phase difference of π is completely equivalent to a phase difference of π+2nπ for any integer n.

7. Mar 25, 2017

### Techno_Knight

I see. The manual just gives an answer of 9 rad as well, so I figured that was correct enough. I've never come across this though. We always just wrote down Δφ = "whatever the end result was".

Ah, okay, I think I get it now. Due to each phase φ increasing by n*2π each cycle, the Δφ, which is the difference between the two, would make the result the same again. Yeah, I think I've got it. I'll take a look at the theory again as well.

Thanks a ton for the help!