# Hello,Question.Two people stand beside a long fence. One places

1. May 16, 2010

### Awsom Guy

Hello,
Question.
Two people stand beside a long fence. One places her ear against the fence while the other person gives the fence a sharp knock. Two sounds seperated by 0.7s are heard by the listener. If the speed of sound is 335m/s in the air and 5300m/s in the metal fence, how far apart are the people?

My attempt.
335 * 0.7 = 234.5
5300 * 0.7 = 3710

3710 - 234.5 = 3475.5

They are 3475.5m apart.

is this right, please check
Thanks

2. May 16, 2010

### Stonebridge

Re: Waves

The 1st sound reaches in a time T1 where T1=s/5300
The 2nd sound reaches in a time T2 where T2=s/335
s is the distance required.
So T2 - T1 is 0.7s
From this find s.

3. May 16, 2010

### Awsom Guy

Re: Waves

Ummm.... How does that work.
Is is like a equation.
Sorry I'm confused.
Thanks

4. May 16, 2010

### Stonebridge

Re: Waves

The time difference was 0.7 s according to the question.
time is distance / speed
time (T1) for the sound in air is distance / 335
time (T2) for the sound in the metal was distance / 5300
So T1 = s / 335 and T2 = s / 5300
we know T1 - T2 = 0.7
so 0.7 = s/335 - s/5300
It's over to you to find the value of s, the answer to the question.