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-   -   Speed of sound and sound detection lag between ears (http://www.physicsforums.com/showthread.php?t=182692)

 aliaze1 Sep2-07 04:15 PM

Speed of sound and sound detection lag between ears

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

One cue your hearing system uses to localize a sound (i.e., to tell where a sound is coming from) is the slight difference in the arrival times of the sound at your ears. Your ears are spaced approximately 20 cm apart. Consider a sound source 5.0 m from the center of your head along a line 45 degrees to your right.

What is the difference in arrival times? Give your answer in microseconds.

2. Relevant equations

v=d/t
speed of sound in air @ rm temp = 343m/s

3. The attempt at a solution

I set up the problem as so:

http://photo.ringo.com/230/230995202O179609724.jpg

http://photo.ringo.com/230/230995202O179609724.jpg

http://photo.ringo.com/230/230995202O179609724.jpg

and then calculated the two dark red lines as different hypotenuses of the two different triangles (one is L+0.1 and the other is L-0.1), then I used v=d/t (with some manipulations) and got a difference which was incorrect

then i tried to set it up as 5 being the main hypotenuse (dashed turquoise), and did a similar process, again wrong

 learningphysics Sep2-07 04:22 PM

I believe the 5m should be along the dashed 45 degree line... Use the cosine law to calculate the length of the two sides...

Use the dashed line, 0.1m and the right hypoteneuse triangle... calculate the length of the hypoteneuse...

Then use the dashed line, 0.1m and the left hypoteneuse triangle... calculate the length of the left hypoteneuse...

 rootX Sep2-07 04:23 PM

i have forgotten that way, but try drawing a per. line to the last dark red line from L-0.1 point,
and you make a safe assumption that two red lines are parallel..
kinda that

it's a square, in case you don't know
and, so using Pythagorean theorem I got 4.12286E-4 s,

and using some approximations, i made it a two lines problem, and got something very similar.

(0.2*5)/5 = difference in length

 aliaze1 Sep2-07 04:43 PM

Quote:
 Quote by learningphysics (Post 1416929) I believe the 5m should be along the dashed 45 degree line... Use the cosine law to calculate the length of the two sides... Use the dashed line, 0.1m and the right hypoteneuse triangle... calculate the length of the hypoteneuse... Then use the dashed line, 0.1m and the left hypoteneuse triangle... calculate the length of the left hypoteneuse...
After doing so, since v=d/t and i want t, i need to flip the velocity, hence making it 1/343

 aliaze1 Sep2-07 04:49 PM

...not working...

 rootX Sep2-07 04:52 PM

but I am pretty sure that L is 5 m.

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