# Eric Hutchinson.  A cajón is a box-shaped instrument that is played by slapping the front face (tappa) by the hands or sometimes with brushes, sticks,

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Eric Hutchinson

 A cajón is a box-shaped instrument that is played by slapping the front face (tappa) by the hands or sometimes with brushes, sticks, etc.

 Some cajóns are simple boxes and others contain snare wires or guitar wires. Different sounds can be created depending on where it is hit.

 Created by African slaves who were brought to the coast of Perú in the 18 th century to work.  “Caja” means box in Spanish.

 Slaves were forbidden to play music so they used objects such as shipping crates and boxes to play music so they were easily disguised.

 Air forced into a cavity, pressure increases. After external force stops, the higher-pressure air inside will flow out.

 Air in the port (neck) has mass.  Diameter of the port is related to the mass of air and volume in the chamber  A cajón is a hollow cavity resonator, so to find the frequency, we need to derive the formula for a Helmholtz Resonator.

SIMPLE HARMONIC FREQUENCY  The derivative of position is velocity:

SIMPLE HARMONIC FREQUENCY  The derivative of velocity is acceleration:

SIMPLE HARMONIC FREQUENCY  We can rearrange:

SIMPLE HARMONIC FREQUENCY  Now substitute with our original position:

SIMPLE HARMONIC FREQUENCY  Let’s review some physics:

SIMPLE HARMONIC FREQUENCY  Now substitute in our acceleration:

SIMPLE HARMONIC FREQUENCY  Derive our frequency by making a substitution:

Simple Harmonic Frequency: A piston of mass m, free to move in a cylinder of area S and length L has simple harmonic motion like a mass on a spring. This is the “neck”.

Starting with K is a spring constant for air: and where is the air density, S is the area of the neck, and L is the length of the neck. V is the volume of the cavity, and c is the speed of sound in air.

DERIVING THE FORMULA…  Start by substituting formulas for K & m:

DERIVING THE FORMULA…  Simplify. Now bring out the c and cancel:

SIMPLE HARMONIC FREQUENCY  Your result will be the following:

 c = Speed of sound in air (1125 ft/sec)  S = Surface area of the sound hole (square ft)  V = Volume of inside cavity (cubic ft)  L = length of resonator’s neck

The air immediately outside the end of the neck takes part in acoustic oscillation. This air makes the neck appear to be acoustically somewhat longer than its physical length. This length increase is called an “end correction”

For a cajón, the length of the neck is zero, however we do need to account for the end correction, which will be our total length. So the length of the neck is:

ALMOST THERE…  Now substitute into the Helmholtz Resonator formula.  We will also substitute for S:

ALMOST THERE…  Now multiply.

ALMOST THERE…  Now pull terms out of the square root.

ALMOST THERE…  Finally simplify.

ALMOST THERE…  We now have the formula we can use to calculate the frequency of a cavity with a large face and no neck:

 V = 11.875*11.125*19  V = 2510.08 in^3  But wait! This is not the true inside volume because there are objects inside that are taking up space.

 Block 1: V = 1.25*1.25*9.75 = 15.23 in^3  Block 2: V = 1.25*1.25*10.63 = 16.6 in^3  Block 3: V = 1.25*1.25*10.94 = 17.09 in^3  Block 4: V = 1.25*1.25*10.44 = 16.31 in^3

 Rest Block:  V = 1*0.25*11 = 2.75 in^3  Dowel Support Blocks:  2.5*0.5*1.94 = 2.42 in^3  22.56*0.5*1.94 = 2.48 in^3

 Dowel:  V = π*(0.63)^2*10.13 = 12.4 in^3  Pivot Pieces:  V = 7.63*0.5*2.38 = 9.05 in^3  V = 6*0.5*2.13 = 6.38 in^3

 Total volume of inside objects is 100.71 cubic inches.  True Interior Volume: 2510.08 – 100.71 = 2409.37 cubic inches.  Helmholtz formula requires this figure to be in cubic feet, so divide 2409.37 by 12^3 to get 1.39 cubic feet. This is our V.

 Sound hole diameter is 4.75 inches  Sound hole radius is 2.375 inches  Converted to feet, this is 0.198, so this is R.

CALCULATE THE FREQUENCY  Plug in our R and V:  Close to F#2 chord

 Hit the tappa with a rubber mallet  Have a microphone record the sound and have a computer display sound wave  Graph measures pressure versus time  Special thanks to the CSN Physics Dept! (Dr. Carlos Delgado & Ted Bellows)

 The period was found by measuring trough to trough. Notice that one cycle is an “M” shape.  Period is 0.008 sec.

 Frequency is defined as 1 / period.  Frequency = 1 / 0.008  Actual Measured frequency is 125 HZ (between B2 and C3 chords)  What could account for the difference?  QUESTION: What happens to the frequency if we reduce the volume of the cajón?

 Now flip and multiply.  Substitute V/2 for V:

 So if you half the volume then the new frequency is 1.4142135623… times the old.  Can we test this in real life?  Simplify:

 Tried to build this cajón to be close to half the volume of the large cajón  All wood is the same (Baltic birch) and same screws/glue  Same size sound hole  Removable snare wires instead of adjustable  This cajón definitely has a higher pitch

 V = 11.875*9.5625*11.1875 = 1270.39 in^2  Must subtract out support blocks. Each one has a volume of 11.875*1.5*1.4375=25.61 in^2  True volume is approximately 1168 cubic inches, which is close to half the volume of large cajón.  1168/12^3 = 0.68 ft^3.

 Sound hole diameter is 4.75 inches  Sound hole radius is 2.375 inches  Converted to feet, this is 0.198, so this is R.

CALCULATE THE FREQUENCY  Plug in our R and V:  Close to C3 chord

 The period was found by measuring the cycle indicated on the picture.  Period is 0.0059 sec.  Frequency = 1/0.0059 which is 170 Hz. (F3)

 Measured frequency of large cajón: 125 Hz  Measured frequency of small cajón: 170 Hz  125*√2 = 176.77 HZ  So the experimented frequencies are close to a square root of 2 difference, which the math predicts!

 Now take out the 2 from the radical.  Substitute R/2 for R:

 So if you half the radius then we expect the new frequency to be 0.70710678118… times the old.  Rationalize.

 I will run my hand across the back of the cajón as I hit it.  As I move my hand you should hear a change in pitch of the tapping.  Do you hear the difference?  Is the frequency higher or lower?

 Eric Hutchinson  College of Southern Nevada  eric.hutchinson@csn.edu eric.hutchinson@csn.edu REFERENCES:  Kicak, Peter. "FREQUENCY AND DYNAMICS ANALYSIS OF BASS TONE OF CAJON BOX DRUM." Acoustics. Electotechnics and Applied Mechanics, n.d. Web. 16 May 2013.  Fletcher, Neville. 1998. The Physics of Musical Instruments. New York: Springer-Verlag.  Ingard, Uno. 1953. “ON THE THEORY AND DESIGN OF ACOUSTIC RESONATORS”. Journal of the Acoustical Society of America. 25(6).

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