# Length of a Closed Air Column

1. Jan 15, 2016

### hamza2095

1. The problem statement, all variables and given/known data
The resonant length of a closed air column at the first resonance is 0.375m, what is the wavelength when at 4th reasonance

2. Relevant equations

3. The attempt at a solution
1/4λ=0.375m
λ=1.5m
1.5m/7 = 0.214m

2. Jan 15, 2016

### Merlin3189

3. Jan 15, 2016

### TSny

4. Jan 15, 2016

### hamza2095

This was a question on a test, the way I got the answer was knowing v/f=λ and i knew that 7/4 of the wave fits in the 4th reasonance and 1/4 in the first, meaning the frequency is 7x more. I assumed the the velocity was 1.5m/s and the frequency was 1hz so 1.5ms/1hz = 1.5m, after that i mutliplied the 1 hz by 7 to get the answer. Is there an easier way to go about this?, I had trouble with it because the neither the speed of sound or frequency was given.

5. Jan 15, 2016

### TSny

That's the key.

Note that the 0.375 m given in the problem represents the length of the pipe. So 7/4 λ = 0.375 m.

You don't need to assume a value for f.

6. Jan 15, 2016

### hamza2095

Ohhh okay I see, thank you. I knew there was an easier way to look at it.

7. Jan 15, 2016

### TSny

Good. Drawing pictures for these types of problems is always a good thing to do.

8. Jan 16, 2016

### Merlin3189

Absolutely agree with TSny, draw diagrams. I did.

You may notice my first post is oddly worded, because you tricked me into thinking you'd done the second part wrongly and I had to edit it!
I had $\frac{λ_1}{4}=0.375$ and $\frac{7λ_2}{4}=0.375$ so when I saw your /7 I thought at first you'd missed out the 4.