Wave Problem

  • Thread starter fredrick08
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  • #1
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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 doesnt seem right????
what about the x,y,z co-ords, y tell me them?????? plz does anyone have any idea??
 

Answers and Replies

  • #2
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srry bout all them, pi's they are meant to be multiplied, for some reason they look superscript, dont quite know how to use this equation editor yet....
 
  • #3
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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, doesnt seem quite right.....
 
  • #4
alphysicist
Homework Helper
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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
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oh ok, yer thats what i was thinking, but how do i find out the distance from each other in 3d??? i dont 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
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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
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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
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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
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wow kool, didnt know that lol, never worked in 3d before...... lol, anyways thx so much, ur a legend!!!!!! lol
 
  • #10
alphysicist
Homework Helper
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Sure, glad to help!
 

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