A pan flute is based on standing wave resonance in a tube that is open where the
musician blows and closed at the bottom end with bee’s wax.
(a) Describe the standing wave pattern of air displacement in such a tube when a tone
is played.
(b) Determine the length of the tube needed to produce a 440 Hz fundamental
frequency.
(c) Determine the sequence of harmonic frequencies that are supported as standing
waves in that tube.
(d) A given tube is found to be slightly out of tune producing a frequency of 445 Hz
rather than 440 Hz. By what distance should the bee’s wax be moved and in what
direction?
(e) Will the pan flute sound out of tune playing alone on a cold day? What about if
the pan flute plays along with an electronic piano on a cold day?
(f) What happens to the frequency and the set of harmonics of the 440 Hz tube if the
bee’s wax were to completely fall out of a tube so that the bottom of the tube is
now open.
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(a) When a tone is played on a pan flute, the standing wave pattern of air displacement in the tube is determined by the fundamental frequency and its harmonics. The tube acts as a resonator, meaning it amplifies certain frequencies while dampening others.
In a standing wave pattern, nodes and antinodes are formed. A node is a point of minimum displacement where the air molecules do not move, while an antinode is a point of maximum displacement where the air molecules oscillate with maximum amplitude. In a closed-end tube like the one described, a node is formed at the bottom end of the tube, where it is closed with bee's wax, and an antinode is formed at the open end where the musician blows. The air inside the tube oscillates back and forth, creating a standing wave pattern.
(b) To determine the length of the tube needed to produce a fundamental frequency of 440 Hz, we can use the formula for the fundamental frequency of a closed-end tube:
f = v / (2L)
Where f is the fundamental frequency, v is the speed of sound, and L is the length of the tube. Rearranging the formula, we have:
L = v / (2f)
Given that the fundamental frequency is 440 Hz and the speed of sound is approximately 343 m/s at room temperature, we can substitute these values into the formula:
L = 343 m/s / (2 * 440 Hz)
Calculating the result, the length of the tube needed to produce a 440 Hz fundamental frequency is approximately 0.390 meters.
(c) The sequence of harmonic frequencies that are supported as standing waves in the tube are integer multiples of the fundamental frequency. In other words, the supported harmonics will be multiples of 440 Hz. The second harmonic will be 2 * 440 Hz = 880 Hz, the third harmonic will be 3 * 440 Hz = 1320 Hz, and so on.
(d) If the given tube is producing a frequency of 445 Hz instead of 440 Hz, it means it is slightly sharp. To correct this, the length of the tube needs to be increased. Moving the bee's wax in the direction towards the open end of the tube will increase the effective length of the tube, thereby lowering the frequency. The bee's wax needs to be moved a distance corresponding to an increase of 5 Hz in frequency.
(e) The pan flute may sound slightly out of tune on a cold day. This is because the speed of sound in air is temperature-dependent. As the temperature decreases, the speed of sound decreases, resulting in a lower frequency for a given length of the tube. However, the effect of temperature on the pan flute's tuning is relatively small and may not be noticeable to the average listener.
If the pan flute plays along with an electronic piano on a cold day, there may be a noticeable difference in tuning. Electronic pianos are designed to produce accurate pitches based on standard tuning systems. If the pan flute is not properly tuned, it may sound noticeably out of tune when played alongside the electronic piano.
(f) If the bee's wax were to completely fall out of the tube, resulting in the bottom of the tube being open, the tube would no longer act as a closed-end resonator. In this case, the fundamental frequency and the set of harmonics would change. The tube would now support a different set of harmonics, known as an open pipe or open-end resonator.
For an open pipe, the fundamental frequency is given by:
f = v / (2L)
Where f is the fundamental frequency, v is the speed of sound, and L is the length of the tube. With the bottom of the tube open, the length is effectively doubled:
L = 2 * L'
Replacing L in the formula, we get:
f = v / (2 * L')
Thus, if the pan flute with a length of L produced a fundamental frequency of 440 Hz when closed at the bottom, when the bottom is open, the new fundamental frequency would be halved, resulting in a frequency of 220 Hz. The set of harmonics would also change accordingly, with the second harmonic at 440 Hz, the third harmonic at 660 Hz, and so on.