A pan flute is based on standing wave resonance in a tube that is open where the

Question

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. Describe the standing wave pattern of air displacement in such a tube when a tone is played. Determine the length of the tube needed to produce a 440 Hz fundamental frequency. Determine the sequence of harmonic frequencies that are supported as standing waves in that tube. A given tube is found to be slightly out of tunic producing a frequency of 445 Hz rather than 440 Hz. By what distance should the bee's wax be moved and in what direction 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 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.

Explanation / Answer

node: point where displacement is zero.

antinode: point where displacement is maximum

part a:

as one end is closed, a node is created at closed end

an anti node exists at open end

the standing wave pattern is generated depending upon frequency with every time

node at closed end and anti node at open end.

part b:

for fundamental frequency, length of the tube=(1/4)*wavelength

as wavelength*frequency=speed of sound

==>wavelength=speed/frequency=343/440=0.77954 m

==>length of the tube=(1/4)*0.77954=0.194885 m

part c:

to satisfy node at closed end and anti node at open end,

length of the tube can be (1/4) times, (3/4) times, (5/4) times ...wwavelength.

hence frequency will be odd integral multiple of fundamental frequency

i.e. 440 Hz, 440*3=1320 Hz, 440*5=2200 Hz etc

part d:

for frequency 445 Hz, wavelength=speed/frequency=343/445=0.770786 m

length of tube required=(1/4)*wavelength=0.192697 m

then the bee wax is to be moved inwards and by a distance of 0.194885-0.192697=2.188*10^(-3) m

e)speed of sound varies with temperature. so if it is tuned for a particular freqeuncy, and speed of sound is changed,

then frequency will be altered and it will sound out of tune.

f)for open ended tube, both end are anti nodes.

then length of tube=(1/2)*wavelength for fundamental frequency

==>wavelength=2*length of the tube=0.38977 m

then fundamental frequency=speed/wavelength=343/0.38977=880 Hz

as length of the tube will be equal to integral multiples of (1/2)*wavelength, frequency values will be

integral multiple values of fundamental frequency

i.e. 880 Hz,1760 Hz,2640 Hz etc.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.