Speed of sound and sound detection lag between ears

AI Thread Summary
The discussion focuses on calculating the difference in sound arrival times at the ears, given a sound source located 5 meters away at a 45-degree angle. The speed of sound in air is noted as 343 m/s, and the problem involves using the Pythagorean theorem and cosine law to determine the distances from the sound source to each ear. Various attempts to solve the problem are shared, including setting up triangles and manipulating equations, but the calculations have led to incorrect results. The importance of accurately representing the geometry of the situation is emphasized, particularly regarding the angles and distances involved. Ultimately, the goal is to find the time difference in microseconds for sound detection between the ears.
aliaze1
Messages
173
Reaction score
1

Homework Statement



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.

Homework Equations



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

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
 
Last edited by a moderator:
Physics news on Phys.org
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...
 
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
 
Last edited:
learningphysics said:
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
 
...not working...
 
but I am pretty sure that L is 5 m.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top