Eric Hutchinson. A cajón is a box-shaped instrument that is played by slapping the front face...

THE MATHEMATICS OF ACAJÓN DRUM

Eric Hutchinson

WHAT IS A CAJÓN?

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.

WHAT IS A CAJÓN?

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

HISTORY OF THE CAJÓN

Created by African slaves who were brought to the coast of Perú in the 18th century to work.

“Caja” means box in Spanish.

HISTORY OF THE CAJÓN

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

HELMHOLTZ RESONATOR

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

HELMHOLTZ RESONATOR

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

We start with the position:

tAy sin


The derivative of position is velocity:

)cos( tAdt

dyV


The derivative of velocity is acceleration:

tAdt

dVa sin2


We can rearrange:

)sin(2 tAa


Now substitute with our original position:

ya 2


Let’s review some physics:

Kymaso

MotionofLawSecondsNewtonmaF

LawsHookeKyF

)'(

)'(


Now substitute in our acceleration:

m

K

Km

Kyym

2

2


Derive our frequency by making a substitution:

m

Kfsom

K

f

FrequencyandPeriod

2

1

2

2

2

DERIVING THE FORMULA…

Simple Harmonic Frequency:

A piston of mass m, free to movein a cylinder of area S and length Lhas simple harmonic motion like amass on a spring. This is the “neck”.

m

Kf

21


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.

m

Kf

21

V

cSK

22

SLm


Start by substituting formulas for K & m:

SLVcS

m

Kf

22

2

1

2

1


Simplify. Now bring out the c and cancel:

SLV

cSf

1

2

1 22


Your result will be the following:

VL

Scf

2

HELMHOLTZ RESONATOR FORMULA

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

VL

Scf

2

LENGTH OF THE “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”

LENGTH OF THE “NECK”

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: RL 316

ALMOST THERE…

Now substitute into the Helmholtz Resonator formula. We will also substitute for S:2R

RL 316

RV

Rc

VL

Scf

31622

2

ALMOST THERE…

Now multiply.

16

3

2

2

VR

Rcf

ALMOST THERE…

Now pull terms out of the square root.

V

Rcf

16

3

2

2

ALMOST THERE…

Finally simplify.

V

Rcf

3

42

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

Rcf

3

8

CALCULATE THE VOLUME

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.

VOLUME OF SUPPORT BLOCKS

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

VOLUME OF SNARE ASSEMBLY

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

VOLUME OF SNARE ASSEMBLY

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

TRUE INTERIOR VOLUME

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 RADIUS

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 HZf

f

V

Rcf

92

39.1

)198.0(3

8

1125

3

8

REAL WORLD EXPERIMENT

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)

GRAPH OF PRESSURE VS. TIME

GRAPH ANALYZING

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

Period is 0.008 sec.

THE MEASURED FREQUENCY IS…

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?

HALVING THE VOLUME…

Now flip and multiply.

Substitute V/2 for V:

2

3

8

3

8

VRc

f

V

Rcf

HALVING THE VOLUME…

So if you half the volume then the new frequency is 1.4142135623… times the old.

Can we test this in real life?

Simplify:

V

Rcf

V

Rcf

3

82

32

4

INTRODUCE THE SMALL CAJÓN!

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

SMALL CAJÓN : VOLUME 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 RADIUS

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 HZf

f

V

Rcf

131

68.0

)198.0(3

8

1125

3

8

GRAPH OF PRESSURE VS. TIME

GRAPH ANALYZING

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)

WHAT IS THE SIGNIFICANCE?

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!

HALVING THE RADIUS…

Now take out the 2 from the radical.

Substitute R/2 for R:

V

Rcf

V

Rcf

2

3

8

3

8

HALVING THE RADIUS…

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

Rationalize.

V

Rcf

V

Rcf

3

82

2

3

2

1

8

LISTEN CLOSELY!

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?

THANK YOU!

Eric Hutchinson College of Southern Nevada [email protected]

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. <34iac.acoustics.sk/proceedings/Kicak.pdf>

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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Eric Hutchinson. A cajón is a box-shaped instrument that is played by slapping the front face...

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