Find fundamental freq. and 1st overtone closed each end pipe

[moenste]

Homework Statement

Calculate the frequency of: (a) the fundamental, (b) the first overtone, produced by a pipe of length 40 cm which is closed at each end. (Velocity of sound in air = 340 m s^-1.)

Answers: (a) 425 Hz, (b) 850 Hz

2. The attempt at a solution
L = 0.4 m, v = 340 m s^-1

(a) f₁ = ?
f_n = nv / 2L
f₁ = 1 * 340 / 2 * 0.4 = 425 Hz

(b) f₂ = ?
f₂ = 2f₁
f₂ = 2 * 425 = 850 Hz

The answer fits but I used the formulas which are for open pipes. When I used the formulas for the closed pipes I got wrong answers (212.5 Hz and 637.5 Hz). Why "closed pipe at each end" is considered an open pipe? Or maybe I used the formulas wrong and the solition is different?

 

[TSny]

To see why the case of "closed at each end" gives the same frequencies as the case "open at each end", you should draw several standing wave patterns for the two cases.

What is the wavelength for the fundamental mode in each case?

What is the wavelength for the next overtone in each case?

 

[moenste]

To see why the case of "closed at each end" gives the same frequencies as the case "open at each end", you should draw several standing wave patterns for the two cases.

What is the wavelength for the fundamental mode in each case?

What is the wavelength for the next overtone in each case?

The closed-closed pipes look like the open-open and the only difference is that it's like they were moved to the left or right. Also n = 1, 2, 3, ... and not n = 1, 3, 5, ... as for the open-closed pipe.

So, it looks like we need to use the formulas for an open pipe for a closed-closed pipe. And also threat it as an open-open pipe?

 

[TSny]

Yes.

$f = v/\lambda$ where $v$ is the fixed speed of sound. So, the frequencies are determined by the wavelengths. As you noted, the wave patterns for the open-open and closed-closed differ by just an overall shift to the left or right. Another way to look at it is that in going from closed-closed to open-open you just interchange nodes and antinodes. So, the distance between consecutive nodes and antinodes ($\lambda/4$) is the same for each case.

 

[moenste]

Another way to look at it is that in going from closed-closed to open-open you just interchange nodes and antinodes. So, the distance between consecutive nodes and antinodes (λ/4\lambda/4) is the same for each case.

I actually though of that in the beginning. That if both parts are closed so the nodes and antinodes should just inverse compared to the open-open pipe.

Thank you :).