Wave Problem

  • Thread starter fredrick08
  • Start date
  • #1
fredrick08
376
0

Homework Statement


a spherical sound wave source at the origin emits a sound wave with frequency 13100Hz and a wave speed of 346m/s. what is the phase difference in degrees and radians between the two points with (x,y,z) co-ords (1cm,3cm,2cm) and (-1cm,1.5cm,2.5cm)??


Homework Equations


v=f[tex]\lambda[/tex]
[tex]\Delta\phi[/tex]=2[tex]\pi[/tex][tex]\Delta[/tex]x/[tex]\lambda[/tex]

The Attempt at a Solution


[tex]\lambda[/tex]=v/f=346/13100=26.4x10^-3m
[tex]\Delta[/tex][tex]\phi[/tex]=2[tex]\pi[/tex](.01--.01)/[tex]\lambda[/tex]=
2[tex]\pi[/tex](.02)/26.4x10^-3=4.76rad=272.6degrees

i know this question seems simple but is it really just that, it doesn't seem right?
what about the x,y,z co-ords, y tell me them? please does anyone have any idea??
 

Answers and Replies

  • #2
fredrick08
376
0
srry bout all them, pi's they are meant to be multiplied, for some reason they look superscript, don't quite know how to use this equation editor yet...
 
  • #3
fredrick08
376
0
it doenst help that there is not an example of a question like this in my book, just the formula... but 4.76rad and 272.6degrees, doesn't seem quite right...
 
  • #4
alphysicist
Homework Helper
2,238
2
Hi fredrick08,

Homework Statement


a spherical sound wave source at the origin emits a sound wave with frequency 13100Hz and a wave speed of 346m/s. what is the phase difference in degrees and radians between the two points with (x,y,z) co-ords (1cm,3cm,2cm) and (-1cm,1.5cm,2.5cm)??


Homework Equations


v=f[tex]\lambda[/tex]
[tex]\Delta\phi[/tex]=2[tex]\pi[/tex][tex]\Delta[/tex]x/[tex]\lambda[/tex]

The Attempt at a Solution


[tex]\lambda[/tex]=v/f=346/13100=26.4x10^-3m
[tex]\Delta[/tex][tex]\phi[/tex]=2[tex]\pi[/tex](.01--.01)/[tex]\lambda[/tex]=


I don't think this is right. The important thing for the phase changes is the distance the waves are for the source. If this was a one-dimensional wave in the x-direction, you would just subtract the x-coordinates. What would you need to do for a three-dimensional case? Once you have that, I think the rest is straightforward.
 
  • #5
fredrick08
376
0
oh ok, yer that's what i was thinking, but how do i find out the distance from each other in 3d? i don't quite understand how to draw a 3d graph... the difference between the x,y,z is (0.02,-0.015,0.005)m...
 
  • #6
alphysicist
Homework Helper
2,238
2
No, what is needed is not the distance between the points, but the difference in how far each point away is from the origin. What do you get?

(For example, if the two points were (1,0,0) and (0,1,0), the phase difference would be zero, because they would be the same distance from the origin.)
 
  • #7
fredrick08
376
0
ok so i got have to do a lot of pythag...

point1=sqrt(0.01^2+0.03^2)=0.031m, and in z dir, sqrt(0.031^2+0.02^2)=0.0374m
point2=sqrt(0.01^2+0.015^2)=0.018m, and in z dir, sqrt(0.018^2+0.025^2)=0.0308m
change=point1-point2=0.374-0.0308=0.0066
put that in the equation and rofl change in phase = .25(2pi)=pi/2 or 90degrees rofl... now that question has been rigged lol, is that right? sounds it lol
 
  • #8
alphysicist
Homework Helper
2,238
2
That looks right to me

You can do the 3-D distance formula in one step, so point 1 would be:

[tex]
d=\sqrt{x^2+y^2+z^2}=\sqrt{0.01^2+0.03^2+0.02^2}
[/tex]
and the same thing for point 2.
 
  • #9
fredrick08
376
0
wow kool, didnt know that lol, never worked in 3d before... lol, anyways thanks so much, ur a legend! lol
 
  • #10
alphysicist
Homework Helper
2,238
2
Sure, glad to help!
 

Suggested for: Wave Problem

  • Last Post
Replies
22
Views
492
  • Last Post
Replies
8
Views
321
  • Last Post
Replies
1
Views
313
  • Last Post
Replies
3
Views
40
  • Last Post
Replies
4
Views
302
Replies
1
Views
323
Replies
10
Views
324
  • Last Post
Replies
2
Views
273
  • Last Post
Replies
4
Views
398
Replies
4
Views
272
Top