Calculating Resonant Frequencies of Closed Air Columns

[qszwdxefc]

Homework Statement

A closed air column is 60.0cm long. Calculate the frequency of the forks that will cause resonance at:

a) the first resonant length
b) the second resonant length

Note that the speed of sound is 344m/s.

Homework Equations

Ln = (2n - 1) * \lambda / 4

f_n = (2n - 1)f₁, where f₁ = V/4L and 4L = \lambda

The Attempt at a Solution

I cannot figure out what the actual magnitude of the resonant length is, and don't know how to begin solving the problem.

Thanks.

 

[LowlyPion]

You have the equations right there don't you?

You have f1 = v/(4λ) looks like to me.

Maybe this link helps?
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/clocol.html#c1

 

[qszwdxefc]

On the back of the page, the question additionally asks for the third resonant length.

The textbook provides answers for a, b, and c as 143Hz, 430Hz, and 770Hz respectively. When I reversed the calculations for c), they used a third resonant length of 60.0cm, which was the length of the whole air column.

I still can't figure out how they got the answers for a and b though :S.

 

[LowlyPion]

On the back of the page, the question additionally asks for the third resonant length.

The textbook provides answers for a, b, and c as 143Hz, 430Hz, and 770Hz respectively. When I reversed the calculations for c), they used a third resonant length of 60.0cm, which was the length of the whole air column.

I still can't figure out how they got the answers for a and b though :S.

Perhaps I misled with my typo, though you wrote it correctly

But to state your equations again:
f1 = v/(4L)
f_n = (2n - 1)*f1

4*L = 4*.6 = 2.4

f1 = 343/2.4 = 143

f2 = 3*f1

f3 = 5*f1

 

[qszwdxefc]

Makes sense, I understand now.

Thanks again :).