A standing wave is established in an organ pipe that is closed at one end. What is located at the open and closed ends of the pipe?

Question

Lance Pierce

Physics

7 February, 11:10

A standing wave is established in an organ pipe that is closed at one end. What is located at the open and closed ends of the pipe?

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Answers (1)

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Glenn Valentine · Answer 1

- displacement antinode is located at the open end

- displacement node is located at the closed end

Explanation:

The cane tubes or other hollow trunk plants constituted the first musical instruments. They made sound blowing at one end. The air contained in the tube went into vibration emitting a sound.

Modern versions of these wind instruments are flutes, trumpets and clarinets, all developed so that the interpreter produces many notes within a wide range of acoustic frequencies.

The organ is an instrument formed by many tubes in which each tube gives a single note. The organ of the concert hall of the Sydney Opera House completed in 1979 has 10500 tubes controlled by the mechanical action of 5 keyboards and a pedal board.

The organ tube is excited by the air entering the lower end. The air is transformed into a jet in the slit between the soul (a plate transverse to the tube) and the lower lip. The air jet interacts with the column of air contained in the tube. The waves that propagate along the turbulent current maintain a uniform oscillation in the air column making the tube sound.

If a tube is open, the air vibrates with its maximum amplitude at the ends. In the figure, the first three vibration modes are represented

As the distance between two nodes or antinode is half wavelength. If the length of the tube is L, we have to

L = λ / 2, L = λ, L = 3λ / 2, ... in general L = nλ / 2, n = 1, 2, 3 ... is an integer

Considering that λ = Vs / f (speed of sound divided by frequency)

The frequencies of the different vibration modes respond to the formula

f = (n/2) * (Vs/L) with n = 1, 2,3 ...

If the tube is closed, a belly originates at the end through which the air penetrates and a node at the closed end. As the distance between a belly and a consecutive node is λ / 4. The length L of the tube is in the figures represented is L = λ / 4, L = 3λ / 4, L = 5λ / 4 ...

In general L = (2n + 1) λ / 4; with n = 0, 1, 2, 3, ...

The frequencies of the different vibration modes respond to the formula:

f = [ (2n + 1) * Vs] / (4*L) with n = 1,2,3 ...