# Homework Help: Help on Sound problem

1. Oct 11, 2005

### Ryan231

Ive been trying this one for a little while now and keep getting myself stumped here... So the question is...

A person sees a heavy stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in air and the other in the concrete, and they are 1.4 s apart. How far away did the impact occur? (the speed of sound in concrete is about 4000 m/s)

So far I've got...
VConcrete = 4000m/s
VAir = 349 m/s

I've come up with a model that basically has 3 variables...
D1 = total distance
D2 = distance from rock hitting ground to 1.4s mark
D3 = 1.4s mark to person

Ive found that
Dconcrete = 4000*1.4 = 5600m
Dair = 349*1.4 = 488.6m

So I need to find D2 and D3 Lengths using the above somehow , D3 takes 1.4s for the slower sound, which would be the sound moving through the air.

d3 = 1.4s D2 =?
Person l______________l_________Rock
l_______________________l
D1 = total disance

My diagram got screwed by autoformatting.. everything is shifted to the right

2. Oct 11, 2005

### Ryan231

help anyone?

3. Oct 11, 2005

### Skippy

I don't understand what you did. "distance from rock hitting ground to 1.4s mark".

And Dconcrete = 4000 * 1.4s seems wrong as well. 1.4s is the interval between the two sound waves and not the time it takes for the Wave to travel through the concrete.

Actually, I think there's something wrong with the problem itself. Did you write everything correctly?... not miliseconds instead of seconds? Recheck 4000 m/s for the concrete wave too.

4. Oct 11, 2005

### Ryan231

OK, the problem is correct.. 4000m/s for concrete is what was given so thats what I'm using... Forget what I put.. I was just brainstorming.. Can you maybe get me started in the right direction?

Maybe switch D2 with T2.. As in when the person hears the first sound the other one is at the 1.4s mark... So I need to find how long it took that sound to travel from the rock to the 1.4s mark and then solve by adding the 2 times together and plugging in one of the equations..

Last edited: Oct 11, 2005
5. Oct 11, 2005

### Skippy

Not sure if you can do that or not.

You should just be able to use D= V(c)*T(1) and D= V(s) * T(2) where you know T(2) is just 1.4s longer than T(1).

6. Oct 11, 2005

### Ryan231

so... D2 = V(s) * T2 = 488.6m
D1 = 4000 * T1 = ?

I'm not following?

7. Oct 11, 2005

### Skippy

Both the D are the same, since both waves travel the same distance (from where the rock hit to where you are). T(2) = T(1) + 1.4s. You can't directly solve for anything, but you can use the fact that D(1)=D(2) to combine the equations and then solve for T(1).