Question

A tube which is open at one end and closed at the other end has a length of 0.25 m. assume the speed of sound in the tube and in the surrounding air 350 m/s.

a. Make one drawing of the tube for each of the three resonances with the longest wavelengths possible for the tube (i.e. 3 drawings total). Label the wavelength on each drawing (or a fraction of the wavelength if a full wave is not present), the physical quantity that the wave represents, where this quantity is a node and antinode, and which end of the tube is open.

b. A frequency generator near the tube plays exactly one frequency: 1400 Hz. If there will be any resonances in the tube when the frequency generator plays this frequency, what are the wavelengths and frequencies of the resonant waves in the tube? Why?

Answer 1

a. For this type of sonor tubes its length is a multiple of half-wavelength:

The first 3 resonances:

(1)

(2)

(3)

The drawing represents the position (in time) of the particles from the air column inside the tube.

Node=crossing points of the dotted curve (red arrows in the image)

Antinode="bulges" of the dotted curves (blue arrows in the image)

From the image is clear which end is open (top).

b.

Replace the numbers and obtain: n=2.

Harmonics 1 and 2 are obtained. The formula answers to the question "why?"