The overall length of a piccolo is 27.0 cm. The resonating air column is open at both ends.

Question

(a) Find the frequency of the lowest note a piccolo can sound. (Assume that the speed of sound in air is 343 m/s.)

(b) Opening holes in the side of a piccolo effectively shortens the length of the resonant column. Assume the highest note a piccolo can sound is 5 000 Hz. Find the distance between adjacent antinodes for this mode of vibration.

Answer 1

For standing waves in a column of air in a pipe open at both ends, it can be shown that the frequencies of the waves that can form standing wave patterns is:



fn=nv2l



where v is the speed of sound in air (343 m/s) and l is the length of the air column (the length of the imstrument).



If you insert n=1 you get the fundamental frequency, the lowest tone that can be played on that instrument. (Higher whole-number values of n are called the harmonics.)

 Using the values in this problem, we see that

f1=3432⋅0.32=536 Hz

This is approximately one octave above middle C.

) The frequency of the lowest note a piccolo can play is; f = 535.94 Hz



B) The distance between adjacent antinodes for this mode of vibration is; D = 42.875 cm



Resonance calculations



A) The formula for the frequency of the lowest note a piccolo can play is gotten from the formula;



f = v/2L



Where;



v is speed of sound in air = 343 m/sL is length given as 32 cm = 0.32 m



Thus;



f = 343/(2 * 0.32)



f = 535.94 Hz



B) Formula for wavelength is;



λ = v/f



Thus;



λ = 343/4000



λ = 0.08575 m



Now, the distance between adjacent antinodes for the given mode of vibration is;



D = λ/2



D = 0.08575/2



D = 0.042875 m



This gives; D = 42.875 cm