# Can someone check my answers (Closed and open tubes)

1. Nov 7, 2008

### viciousp

1. The problem statement, all variables and given/known data
A tube, open only at one end, is cut into two shorter (non-equal)
lengths. The piece open at both ends has a fundamental frequency
of 425 Hz, while the piece open only at one end has a fundamental
frequency of 675 Hz. What is the fundamental frequency of the
original tube ?

also
Which of the following describes the relative temperatures in the refraction of sound. N is the normal line.

http://img511.imageshack.us/my.php?image=refractionfe5.jpg

2. Relevant equations
L1+L2=L
V/2L=frequency of open tube
V/4L=Frequency of closed tube

3. The attempt at a solution
For the first question I said the V/2(L1)=425 and V/4(L2)=675. so V=850L1 and V=2700L2. so I solved for L1 and got L1=3.176L2 and using the above equation L=4.176L2. So 2700L2/4(4.176L2)=161.63Hz, it seems resonable but I just want to see if I am right.

As for the other question I said that it moves from cold air to warm air since the angle gets larger and then from warm air to cold air since the angle gets smaller.

2. Nov 7, 2008

### krausr79

Couldn't see the image, but I think this work is good.

3. Nov 7, 2008

### viciousp

Here is the image again

http://img511.imageshack.us/img511/4403/refractionfe5.jpg [Broken]

Last edited by a moderator: May 3, 2017
4. Nov 9, 2008

Anyone?