Reflection and transmission of waves

AI Thread Summary
The discussion revolves around a physics problem involving sound waves and echoes in a river canyon. A participant is confused about the time interval of 0.407 seconds between echoes after clapping hands, questioning how to relate this to the distance from the canyon walls. The speed of sound in air is noted as 343 m/s, and the participant attempts to calculate the distance to the closer wall using the time interval. Clarifications are provided that there are two echoes, one from each wall, and the participant ultimately resolves their confusion. The problem highlights the importance of understanding sound reflection and the geometry of the canyon.
begbeg42
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Homework Statement



A river canyon is 147 m across as shown in Figure P.32. You are located on the river surface and clap your hands once to generate a brief acoustic pulse. You detect a time interval of 0.407 s between the twin echoes recorded on tape, analyzed when back home (?) in a physics lab. Where were you relative to the canyon walls? (Neglect multiple echoes.)

Answer is m (from the closer wall)


Homework Equations



Not sure

The Attempt at a Solution


not sure
 
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Hi begbeg42! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
is this problem anything like where you drop a rock into a well and hear the sound ___ sec. later and have to find the depth? I know how to do this...but this problem confuses me b/c it involves horizontal distances...

I know the speed of sound in 20 degree Celsius air is 343 m/s
I'm not sure physically/intuitively what .407 sec. actually means? is it the time it takes you to hear the echo after you clap?

im utterly baffled. and yes I have read about reflections of sound/echoes etc.
sorry...and any guidance is appreciated
 
Horizontal and vertical are the same …

but in this case you have two sounds, the .407 s is the gap between the two echoes :smile:
 
t1(sound to travel to wall) + t1(sound to return to you)=a=2t(sound to travel to wall)
t2(sound to travel to wall) + t2(sound to return to you)=b=2t(sound to travel to wall)
4t(sound to travel to wall)=0.407 sec?
t(sound to travel to wall)=.407/4?

d(to wall)=343(.407/4)=34.9...which seems plausible but

i mean this seems too simple...how does the actual width of the canyon=147 m come into play?
 
Last edited:
Draw a diagram

what is a + b?

what is a - b? :smile:
 
t(echo1) - t(echo2)=0.407
t(echo1)=t(sound to travel to wall) + t(reflected sound to return to you)
t(echo2)=?

d(from wall)=343t(sound to travel to wall)

does the sound hit the wall, return to you, and then bounce back or...how exactly is the second echo produced?
 
begbeg42 said:
does the sound hit the wall, return to you, and then bounce back or...how exactly is the second echo produced?

ah … you're not getting this canyon thing, are you?

there are two walls … one echo comes from each wall.

Try again! :smile:
 
wow...sorry about that

i got it anyways...took me awhile, though I don't know why

thanks anyways!
 
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