16.5 CHAPTER 16. THE PHYSICS OF MUSIC
L 1 = 18,2 cm
L 2 = 50,3 cm
f = 512 Hz
v =?
Remember that:
v = f× λ
We have values for f and so to calculate v, we need to first find λ. You
know that the differencein the length of the air column between two resonances
is half a wavelength.
Step 2 : Calculate the difference in the length of the air column between the two
resonances:
L 2 − L 1 = 32,1 cm
Therefore 32,1 cm =^12 × λ
So,
λ = 2× 32 ,1 cm
= 64,2 cm
= 0,642 m
Step 3 : Now you can substitute into the equation for v to find the speed of sound:
v = f× λ
= 512× 0 , 642
= 328,7 m.s−^1
16.5 Music and Sound Quality ESBHC
In the sound chapter, wereferred to the quality ofsound as its tone. Whatmakes the tone of a note
played on an instrument? When you pluck a string or vibrate air in a tube, you hear mostly the
fundamental frequency.Higher harmonics are present, but are fainter. These are called overtones.
The tone of a note depends on its mixture of overtones. Different instruments have different mixtures
of overtones. This is why the same note soundsdifferent on a flute and apiano.
Let us see how overtones can change the shapeof a wave:
