# The speed of sound

1. Apr 17, 2007

Alright we did a lab using the speed of sound we were suppose to able to calculate the temperature outside.
v = 332 +0.6T

So this is what we did, one person hit two blocks together twenty times at an equal pace. The timers start when the person first hits the block together and stops when the last echo is heard.
The distance from the wall and back is 200m
So the times we got were 23.40s, 23.28s

Then I calculated the average total time and average time for each echo cycle. Which is 1.167s

So Speed = Distance / Time for each cycle Right?
Speed = 200m/1.167s?
If I do this, I end up with a negative temperature.

So then did this, Speed = Distance/ Half of the time for each cycle.
Speed = 200m/0.5835s
By doing this I got the right temperature.

Can anyone explain to me why I needed to divide the time for each cycle by two?

2. Apr 17, 2007

### hage567

If your time is from when you clap to when you hear the echo, the distance is actually 400m since the sound is making a round trip. (Or half the total time to go 200 m)

3. Apr 17, 2007

What do you mean its making a round trip?
Do you mean like its 400m to the wall and back?
Because the distance to the wall is 100m.

4. Apr 17, 2007

### hage567

Hmmm, I misread part of your post, I thought the 200 m was the distance to the wall only. Sorry about that.

5. Apr 17, 2007

Yea its no problem. But do you understand why the time would be divided by two or why the distance would be multiplied by two?

6. Apr 17, 2007

### denverdoc

helluva way to measure temperature if you ask me:tongue:

But something is amiss, 200/1.17=170m/s, exactly 1/2 the speed of sound. I think maybe something in the methods used which caused the receiver to hear an earlier echo than the one triggered by the last event? I would think cleaner, tho less efficient and I understand the reasoning I believe, but led to big error anyway, single trials x 20..

7. Apr 17, 2007

Huh? =/ lol sorry, I still dont understand why its like this.

8. Apr 17, 2007

### hage567

Just to check, are you SURE the distance to the wall and back was 200 m??

I think what denverdoc is saying is that it could be due to error in measurement, like stopping the timer too soon, or maybe just in the reaction time of the person working the timer.

9. Apr 17, 2007

### denverdoc

The notion I suppose is a training effect and cancels sone randomness from the human side, in other words observations 10-20 show less
variability than 1-10. Thats the strength of the experiment. The weakness is hearing slap/echo/slap/echo and avoiding echo/slap error. In plain english after ten trials are the timekepers hearing the sound fom an earlier echo?

Last edited: Apr 17, 2007
10. Apr 17, 2007

Yup I'm sure the distance there and back is 200m.

11. Apr 17, 2007

No I don't think the time keepers are hearing the sound from an earlier echo, it was pretty clear 20 echoes were heard.

12. Apr 18, 2007

### denverdoc

i didn't buy it either. especially if it was just an aggregate time. Hage is right, had to have been 50 meters away. 100m is like a football field.

13. Apr 18, 2007

alright I found out why it is now, by hitting it at an equal pace, there is a pause between each echo and the next hit.

Hit-(speed moves)-Echo-(pause to create equal pace)-Hit-(speed moves)-Echo-(pause to create equal pace)

So the 1.167s is the time between the hits. Therefore I had to divide it by two.

14. Apr 18, 2007

### denverdoc

good grief, why didn't i see that, thanks for the update.