From: S-----------e@aol.com [mangled for privacy]
Date: Wed, 3 Jan 2001 22:48:38 EST
Subject: Urgent! Please Read!
To: rusin@math.niu.edu
Hey!
My name is M----- C---, and I'm a freshman at M----- High School (down in
GA)! I am doing a project for math! (ew) analyzing the relationship between
music and math. I play piano, violin, bass...I know there are TONS of math
applications when it comes to music! However, this project's supposed to
allow us to research something we don't already know about! So, I decided to
further my topic and focus specifically on ear training. (My teacher pushed
me in this direction, blah) I was wondering how much time conductors or
teachers need to spend on ear training, and how the ear is able to learn to
distinguish and measure between notes, like intervals or chords! I was
thinking, maybe, to just focus on the mathematical ratios found in music/used
in music theory? Intervals, the frequency of the pitches, etc. I plan on
actually performing ear training for my presentation with the class, and
possibly allowing them to use math applications (ratios) to help them to
determine the difference in the intervals...?? Any information you can
contribute would be wonderful!!! I came across your site just now and found
it to be EXTREMEMLY useful! There are sooo many great ideas! However, I came
across your site kinda late...my project's due soon! Ack! I began sending out
e-mails last month to various music experts and the such, but received NO
replies! Well, please try to respond ASAP if possible! Especially if you have
any ideas, I hope you can help!!! And thank you for taking the time to read
this!! Any contribution would be greatly appreciated...Thanks again!
Melody
From: Dave Rusin
Date: Wed, 3 Jan 2001 23:25:21 -0600 (CST)
To: S------------e@aol.com
Subject: Re: Urgent! Please Read!
I'm not really sure what to tell you. As far as I know, ear training
is just that -- training, not analysis. One learns to hear with more
precision simply by practice. No math need be used.
On the other hand, I guess it's worth noting that in fact one need NOT
train one's ear with as much precision as one might think in some
circumstances. In particular I observe that one need NOT have a
particularly good ear to tune instruments. There is a phenomenon known
as "beats" which can be used to match notes perfectly even if you've
got a pretty untrained ear. Near-bozos can tune a piano this way!
I'll try to explain the math involved (trigonometry). The key here is that
a "note" played by an instrument is a complex set of motions of air molecules,
but that the reason it sounds like a note instead of noise is that the
bulk of the motion of the air molecules can be accounted for as a
back-and-forth vibration in which the air molecules are pressing
towards- and away-from the source of the sound in a regular pattern;
the air pressure at a point varies in time t as a function of the form
p(t) = A sin( Bt+C )
for some constants A, B, and C. Here C is more or less irrelevant,
but A is proportional to the _volume_ (loudness) of the note being played
and B is proportional to the _frequency_ of the note. The tuning fork
you use to tune instruments might be the "A" above "middle C", usually
known an "A440" because the frequency is 440 Hz -- the pressure
cycles between outward and inward and back to outward again 440 times per
second; in this case B is 2*pi*440 (about 2764.6) if t is measured
in seconds.
So here's the mathematical secret to tuning: Suppose we had to tune an
instrument -- a horn, say -- which is supposed to be playing that A.
Suppose we sort of thought it might be just a little sharp. Our task is to
play the note and recognize the sound as being principally the pressure
function I wrote above:
p(t) = A sin( B t + C )
and to decide whether or not the constant B is exactly the desired 2764.6
or a little more (if our horn is sharp) or a little less (if our horn
is flat). You'd think it would take a good ear to hear this; you'd
think you'd need a good ear to listen to a tone and have to estimate
the value of the constant B.
Turns out you don't need to do so at all, and the trick is some mathematics.
Suppose that you played that note on your horn, and the AT THE SAME TIME,
someone stuck that tuning fork. What do you here? Two things are now
making the air molecules jiggle. Your horn is making the pressure change
in time according to this formula
p(t) = A sin( B t + C )
and the tuning fork is trying to change the pressure further, adding a
similar change of the form
p2(t) = A2 sin( B2 t + C2 )
for some new constants A2, B2, and C2. This time we know exactly what B2
is (that 2764.6 thing). The size of A2 depends on how hard we hit the
tuning fork. Anyway, the net effect is that the air pressure right next
to your eardrum is increasing and decreasing in time according to the
more complicated function
f(t) = A sin( B t + C ) + A2 sin( B2 t + C2 )
The beauty of this, mathematically, is that we have some facts from
trigonometry which let us restate this. The sine function happens to
obey the general law that
sin( x + y ) = sin(x) * cos(y) + cos(x) * sin(y)
for any two numbers x and y. In particular (replacing y by -y) we
deduce that it also makes
sin( x - y ) = sin(x) * cos(y) - cos(x) * sin(y)
true for all x and y. Add these to discover that
sin( x + y ) + sin( x - y ) = 2 * sin(x) * cos(y)
for any two numbers x and y. In particular, give me any two numbers
a and b and I can let x = (a+b)/2 and y = (a-b)/2 to see that
sin( a ) + sin( b ) = 2 * sin( (a+b)/2 ) * cos( (a-b)/2 ).
That's almost what we have in the functions f(t) above. In fact, it'd
be exactly what you see in f(t) if the numbers A and A2 were equal.
Turns out I can get slightly more complicated formulas to handle the
general case, but instead let me just ask you to strike that tuning fork
just hard enough to make it exactly as loud as your horn; then A2
will be the same as A and that pressure function will be
f(t) = A * ( sin( B t + C ) + sin( B2 t + C2 ) )
which, using our trigonometric formula above, becomes
A * ( 2 * sin( (Bt+C + B2 t + C2)/2 ) * cos( (B t + C - B2 t - C2)/2 ) )
Now, I know this is an eyeful, so let me point out to you what you're
supposed to see. There are three factors, which I'll take out of order.
In the front, you see (2*A); in the back, you see this:
cos( B3 t + C3 )
where for brevity I've written B3 instead of (B - B2)/2 and C3 for
(C-C2)/2. Finally, in the middle you see
sin( B4 t + C4 )
where in the same way I write B4 for (B + B2)/2 and C4 for (C + C2)/2.
So what is this? Well, it has the same general form as our original
pressure functions: we've written the combined sound of the horn and the
tuning hammer in the form
f(t) = A4 sin( B4 t + C4 )
so that it ought to sound like a single note being played with frequency B4.
Well, look back at the formula for B4: it's the average of B (the
frequency of the note your horn is playing) and B2 (the frequency of the
note the tuning hammer is playing). If you were at all on target with your
horn, then B and B2 ought to be pretty close, and their average, B4,
is in the middle between them. If your horn is just a little sharp of
the right note, the combined sound will be also a little sharp, but only
half as far off from being correct.
Here's the fun part, though. So the combined sound sort of sounds like
a note midway between the horn's note and the tuning fork's note; big deal.
The fun question is, how loud is it? That's the quantity A4 in front
of the last formula, remember. Well, guess what: I cheated a little. This
A4 isn't a number at all, but it's all the leftover stuff from some of
our previous equations: it comes out to
A4 = (2*A) * cos( B3 t + C3 )
As you can see, there are still some t's in here, so that this A4 isn't
constant at all but rather changes in time. Well, it _usually_ changes in
time: notice that if B3 happens to be zero, then there really aren't
any t's in here at all, so that A4 really will be a constant. Now, what
would that mean? Let's see, if B3 = 0 then ... now, what what B3? Oh yeah:
B3 = (B - B2)/2. Well, that's zero precisely when B = B2: in other words,
precisely when B and B2 are the same frequency. So if your horn is
_exactly_ in pitch with the tuning fork, then this A4 is just some
number, and the combined effect of playing the horn and the tuning fork
together is to hear a single note of exactly the same pitch. That's not
really exciting, I guess: that's what you might expect if the two are in tune.
But what if they're not quite in tune? What if the tuning hammer is
perfectly vibrating at 440Hz and the horn is making the air vibrate at
442Hz -- just half a percent sharp? Well, as we noted above, the combined
sound just sounds like a note at the midway frequency of 441Hz -- so
close to true that most of us wouldn't hear anything except another
"perfect A". But how loud is it? The multiplier A4 would be this
A4 = (2*A) * cos( B3 t + C3 )
again, where here the coefficient B3 is NOT zero; it's actually
(2*pi*442 - 2*pi*440)/2 = 2*pi*1. That means A4 is a quantity which goes
up and down over time, between the values of 2*A and -2*A and back to
2*A, taking 1 second to complete the whole cycle (since 1 = (442 - 440)/2.)
Do you see what that means? The loudness will go from loud to zero to
"negative loud" to zero to loud again, once every second. (Don't worry
about "negative loud" sounds; to our ear they're the same as "loud".)
So if your horn is just a little sharp, what will happen when you play it
and the tuning fork together is that you will continue to hear the sound
of what you think is a "perfect A", except that it will warble in and out
oooOOOoooOOOoooOOO... with the loud moments half a second apart. The
key observation is that YOU CAN HEAR THIS. It would take an incredibly
good ear to distinguish 442 Hz from 440Hz, but any idiot can hear a
do-wah sound which is going loud and soft about as fast as you can count
ONE one THOU-sand TWO one THOU-sand THREE...
The do-wah that you're hearing is called the "beat frequency", and the key
observation is that it is just exactly the difference between the frequencies
of the two sounds being played together. Is your horn too sharp? Flatten it
a little and try the experiment again: the do-wah should slow down. Keep
flattening it bit by bit untill the "beats" sound like
ooooooooooooooooooooooooOOOOOOOOOOOOOOOOOOOOOOOOooooooooooooooooooooooooOOO...
and flatten the horn just a tad more until there is no more "beating":
finally you've gotten it so that B2 = B and as we noted before you just
hear the steady, perfect A.
This is a trick regularly used by piano tuners and so on. Using the
so-called circle of fifths you can tune a whole piano by listening for
beats, starting with a single tuning fork for just one string.
Of course you don't need to understand the mathematics in order to
tune the piano, but it's cool to see _why_ exactly this works.
Hope this is the sort of thing you were looking for.
dave
Prof. David Rusin
Director of Undergraduate Studies
Department of Mathematical Sciences
Northern Illinois University
DeKalb, Illinois, 60115 USA
Email rusin@math.niu.edu
Web http://www.math.niu.edu/~rusin/
Telephone 1-815-753-6739
Fax 1-815-753-1112
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