How Does the Wavelength Change at the Fourth Resonance in a Closed Air Column?

[hamza2095]

Homework Statement

The resonant length of a closed air column at the first resonance is 0.375m, what is the wavelength when at 4th reasonance

Homework Equations

The Attempt at a Solution

1/4λ=0.375m
λ=1.5m
1.5m/7 = 0.214m

 

[Merlin3189]

I agree with your wavelength at first resonance and your answer.

 

[TSny]

As third poster, I second your fourth resonance answer.

 

[hamza2095]

As third poster, I second your fourth resonance answer.

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.

 

[TSny]

i knew that 7/4 of the wave fits in the 4th reasonance

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.

 

[hamza2095]

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

 

[TSny]

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

 

[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.