Good morning.
Bobby, what is the equation which we derived in a previous lesson for the harmonic frequencies
of stringed instruments?
♪flipping physics♪
- The frequencies created by a stringed instrument equal n, the harmonic number which equals
positive integers times the quantity speed of the wave in the string divided by 2 times
the length of the string.
- And remember when the harmonic number n = 1, that is called the first harmonic and
the fundamental frequency.
- And all the harmonics are integer multiples of the fundamental frequency.
- Very nice y'all.
Remember, the derivation of this equation is based on the fact that the ends of strings
in strained instruments are fixed, and therefore must be nodes.
This means the standing wave patterns of the first three harmonics look like this...
What we are deriving now are the equations for the harmonic frequencies of wind instruments.
Wind instruments create a frequency and pitch
by creating standing waves in the air columns of the instruments.
Wind instrument examples are trumpets, tubas, clarinets, flutes oboes - you know.
There are two physics categories of wind instruments; there are open pipe wind instruments which
are a pipe which is open at both ends and there are closed pipe wind instruments which
are closed on one end and open on the other.
- What is an instrument which is closed on both ends called?
- It's called a drum.
- Which is not a wind instrument. - Right.
- The two different types of ends have different boundary conditions.
An open end creates an anti-node and a closed end creates a node.
- An open end creates an anti-node and a closed end creates a node.
- Right.
Let's begin with an open pipe instrument like a flute.
Considering both ends are open and an open end is an anti-node, the first standing wave
pattern has two anti-nodes; one at each end and one node right in the middle.
Billy, what do you think the next two standing wave patterns look like?
- Well, I bet we just add one node and one anti-node each time we move to the next standing
wave pattern just like we did for the stringed instruments.
Let's see, that would mean for the next standing wave pattern, there would be three anti-nodes,
one on each end and one in the middle and then two nodes, one between each pair of anti-nodes.
For the next standing wave pattern, we have four anti-nodes and three nodes; one anti-node
at each of the open ends and then two anti-nodes which are each located 1/3 of the way from
each end and then of course the three nodes which are all located between each pair of anti-nodes.
- Wait a second, I think this is the same as the stringed instrument standing wave pattern.
- No it's not...
Look at the stringed instrument standing wave patterns...
They are clearly different.
- Okay, so, let me clarify.
So, I mean, the wavelengths of the wave for each standing wave pattern are the same.
For example, the first standing wave patterns both have half a wavelength equal to L, the
length of the string or pipe.
The second patterns both have one full wavelength equal to L and the third patterns both have
one and a half wave lengths equal to L. That's what I meant.
- Oh...
- Does that mean the frequency equation is the same?
- Wow! I'm impressed, yes.
The wavelengths for the standing wave patterns are the same for both open pipe and stringed instruments.
So yes, the equations are the same.
The harmonic number is the same, v is the speed of the wave in the medium...
So, for stringed instruments, that is the speed of the wave on the string and for open
pipe instruments, it is the speed of sound in air.
Now let's move on to closed pipe instruments like a clarinet.
Remember, closed pipe instruments have one closed end and one open end.
That means the closed end is a node and the open end is an anti-node.
How many wavelengths equal the length of the pipe in this first standing wave pattern for
a closed pipe? Bo?
- I think that is just 1/4 of a wavelength, right?
- Correct Bo, this is 1/4 of a wavelength.
Now please determine the frequency of the sound.
- Okay, the wavelength of the wave then equals four times the length of the pipe.
We know speed equals frequency times wavelength, so frequency equals speed divided by wavelength,
so frequency equals speed divided by 4L, but like last time, let's put a 1 in front of
that to indicate that this is the first harmonic and the fundamental frequency.
- Absolutely Bo.
Billy, what about the next possible standing wave pattern?
- Well, there still has to be a node at the closed end and an anti-node at the open end,
so, I guess uh...
- I think you need to add one node and one anti-node.
- Okay, so, then from left to right, it will go anti-node, node, anti-node, node with all
those equally spaced apart.
That means there is uh, 3/4 of a wavelength in the length of the pipe...
- Billy, that is correct so far, please keep going.
- Okay.
Then the wavelength equals 4L/3 and the frequency still equals speed divided by wavelength or
speed divided by 4L divided by 3.
That means the frequency equals 3 times the quantity speed divided by 4L.
- Bobby please analyze the next possible standing wave pattern.
- OK, for the next standing wave pattern, we again add one node and one anti-node.
So again, it alternates from left to right from anti-node, node, anti-node, node, anti-node
and node, and they are still equally spaced apart.
Now we have 5/4 of a wave equal to the length of the pipe, that makes the wave length equal to 4L/5.
The frequency then equals speed divided by 4L/5 or 5 times the fundamental frequency.
And the one Billy did was 3 times the fundamental frequency, so the first three possible harmonics
are the first third and fifth harmonic.
Does that make sense?
Do closed pipes not have even harmonics?
- That would be strange.
- Bobby you are correct; closed pipe instruments do not even harmonics.
The equation for closed pipe instruments is frequency equals m times the speed of sound
in air divided by 4 times the length of the pipe, where "m" is the harmonic number and
as Bobby pointed out, m can only be odd integers because closed pipes do not have even harmonics.
Just so you know, not everybody uses m instead of n to indicate that only odd harmonics are
possible but I do.
And as you pointed out, the harmonics are odd integer multiples of the fundamental frequency.
Okay, I do want to confess a few details which we glossed over when deriving these equations.
First is that standing wave patterns in air columns have both displacement and pressure
nodes and anti-nodes.
A closed-end is a displacement node because the air cannot move through the end, that closed end.
This makes the open end of a pipe a displacement anti-node.
An open end is a pressure node because the open end is open to the atmosphere and therefore
the air at the open end maintains a constant pressure at atmospheric pressure.
This makes the closed end of a pipe a pressure anti-node.
Given that knowledge, which have our illustrations been using, pressure or displacement nodes and anti-nodes?
- If a closed end is a displacement node, then these illustrations have to be showing
displacement nodes and anti-nodes.
But I do not really get what that means to be honest.
Do either of you get it?
- No.
- Not really.
- Okay, so first, yes, all of the illustrations we have used so far are in terms of displacement
nodes and displacement anti-nodes.
To understand displacement versus pressure nodes and antinodes, let's bring back the
demonstration and corresponding synced animation of a longitudinal standing wave pattern.
- Longitudinal because waves in air are longitudinal waves.
- Right.
So this is a demonstration of a spring with a longitudinal standing wave pattern with
two different standing wave pattern animations.
The bottom animation has displacement on the y-axis and the top animation has density on the y-axis.
Actually, you know what?
As much as I prefer real demonstrations, there are times when animations actually make things
a little more clear and I think this is one of those times.
So, instead of the spring demonstration of the longitudinal standing wave pattern, we're
going to change to a highly idealized animation of air molecules in a standing wave pattern in a closed pipe.
Notice we now have pressure on the y-axis for the top standing wave pattern and we still
have a displacement for the y-axis for the bottom standing wave pattern.
Now, look for a moment at the pressure anti-nodes...
Do you see how the air molecules are essentially not displaced at all?
Again, only specifically at the locations of pressure anti-nodes are the air molecules
essentially not moving at all.
In other words, do you see how pressure anti-nodes, our displacement nodes and pressure nodes,
our displacement antinodes?
- Sure, I can see that each pressure node is a displacement anti-node and each pressure
antinode is a displacement node.
That makes sense; their locations are essentially inverses of one another.
- Correct.
You also see how the air molecules are essentially not moving at the closed end and at the open
end, the air molecules have the largest magnitude motion?
- Yeah, that means a displacement node is at the closed end and a displacement anti-node is at the open end.
- But a pressure antinode is at the closed end and a pressure node is at the open end.
- Again, because their locations are essentially inverses of one another.
- One last thing is that directly at the open end of a pipe is not quite where the location
of the pressure node and displacement anti-node are.
The air actually oscillates outside the end of the pipe a little bit.
You can actually see that when I put a soap bubble near the end of the tube with an air
column resonating in it.
As you can see, the bubble is outside the tube and yet it is being deformed by the displacement
anti-node which is at the open end of the resonance tube.
In other words, the length of the oscillating column of air in a wind instrument or the
effective length of the pipe is a little bit longer than the measured length of the pipe.
For a circular cross-section pipe and the correction of 0.6 times the radius of the
pipe needs to be added to the length of the pipe for each open end.
- So, to get the effective length of a pipe for a closed pipe, we add 0.6 times the radius
of the pipe to the length of the pipe, and for an open pipe, we had two times 0.6 times
the radius of the pipe to the length of the pipe because an open pipe has two open ends
and a closed pipe only has one open end.
- Thank you very much for learning with me today I enjoyed learning with you.