# Homework Help: Phase Difference

1. Nov 30, 2008

### Husker70

1. The problem statement, all variables and given/known data
A tuning fork generates sound waves with a frequency of 246Hz. The waves travel in
opposite directions along hallway, are reflected by end walls, and return. The hallway
is 47.0m long and the tuning fork is located 14.0m from one end. What is the phase
difference between the reflected waves when they meet at the tuning fork? The speed of
sound in air is 343 m/s.

2. Relevant equations
delta r = Phi/2pie x Lambda
f = v/Lambda

3. The attempt at a solution
The first thing I did was delta r = r2 - r1 = (33.0m - 14.0m) = 19.0m
then f = v/Lambda or Lambda = v/f = 343m/s / 246Hz = 1.39 units?
using delta r = Phi/2pie x Lambda
19.0m = Phi/2pie x 1.39
I get like 86.5 degrees and it should be 91.3 degrees
Any help would be appriciated.
Kevin
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 30, 2008

### Staff: Mentor

Realize that the sound reflects from the walls and thus makes a round trip.
OK.
I don't understand your equation. Each wavelength of path difference gives you 360 degrees (or 2 pi radians) of phase difference.

3. Nov 30, 2008

### Husker70

Doc,
Thanks but now I don't either

delta r = (66.0m - 28.0m) = 38.0m

Is the phase difference = kd
where k = 2pie/delta r

Thanks,
Kevin

4. Nov 30, 2008

### Staff: Mentor

How many wavelengths does that correspond to?

5. Nov 30, 2008

### Husker70

Not sure but trying to figure out

6. Nov 30, 2008

### Staff: Mentor

You already found the wavelength. Just divide that wavelength into the path difference.

7. Nov 30, 2008

### Husker70

I am confused as I get the wavelength to be 1.39m
from Lambda = v/f

and path difference is 38.0m
Are these the figures that I use
Kevin

8. Nov 30, 2008

### Staff: Mentor

Yes, those are the numbers that you need.

It's the same as this example: Say you have 423 pennies. Since 1 dollar equals 100 pennies, how many dollars do you have? (I suspect you have no problem solving that one.) Now do this: You have a path difference of 38 m. Since 1 wavelength equals 1.39 m, how many wavelengths do you have?

9. Nov 30, 2008

### Husker70

27.3 wavelengths. I got that but not sure where to put that to find
phase difference.
Thanks for the help. My brain isn't working this early
Kevin

10. Nov 30, 2008

### Staff: Mentor

OK, but don't round off until the last step.

Whole numbers of wavelengths don't lead to a phase difference. Only the fraction left over after getting rid of integral numbers contributes to the phase difference.

11. Nov 30, 2008

### Husker70

Thanks but I don't get this. I'm not sure what to do with this number

12. Nov 30, 2008

### Husker70

Doc,
I still don't understand can you give me a clue what to do with these numbers
Thanks,
Kevin

13. Nov 30, 2008

### Staff: Mentor

Answer these questions. If two waves differ by exactly one wavelength, what's their phase difference? What if they differed by exactly 12 wavelength? What if they differed by 1/3 of a wavelength? What about 2.33 wavelengths?

14. Nov 30, 2008

### Husker70

I think that is where my problem understand is. I'm looking some more but
I believe that if they differ one wavelength they are in the same phase as well
as 12. I'm not sure though
Kevin

15. Nov 30, 2008

### Staff: Mentor

Exactly right! If the path difference is an integral number of wavelengths, then there is no phase difference.

What about when there's a fraction of a wavelength difference?

16. Nov 30, 2008

### Husker70

I think that when it is a phase shift of 90 degrees it is pie/2
and 180 is pie
270 is 3pie/2
But I'm not sure how to use those to find any other fraction
Kevin

17. Nov 30, 2008

### Staff: Mentor

In terms of wavelengths, one wavelength = 360 degrees (or 2 pi radians).

So, for example, a 1/4 wavelength path length difference corresponds to 1/4 * 360 = 90 degrees phase difference.

When figuring out the phase difference based on the path length difference, all that matters is the fractional part of the path length difference. If the path difference is 2.4 wavelengths, then all that counts is the 0.4 wavelengths. And 0.4 wavelengths corresponds to a phase difference of 0.4*360 = 144 degrees.

18. Nov 30, 2008

### Husker70

It's starting to make more sense so going back we have 38.0m / 1.39m
=27.338129496 number of waves. So they are out of phase by
.338129496 or 121.7 degrees?
Doesn't match the answer but I just want to understand the concept
Kevin

19. Nov 30, 2008

### Husker70

When I carry the numbers correctly it works out
to be .25364432 or 91.3 degrees
Thanks,
Kevin

20. Nov 30, 2008