Now that we've gone over the basics of this.
Let's do a quick example.
Let's suppose we have an organ pipe
and we can consider an organ pipe as a pipe with two open ends.
At both ends of an organ pipe, you actually do have an opening.
Let's suppose that our organ pipe is one meter long.
We could ask what is the lowest frequency that we could play in the pipe?
Where something that I haven't mentioned is that the frequency that you hear
corresponds with the pitch of the sound.
So a low frequency are those very deep, low sounds,
where high frequency are those very a high, high pitched sounds
maybe towards the right of a piano, for example.
So what is the lowest frequency that we could play on this pipe,
the lowest, deepest sound if it's a 1-meter long pipe
and we know the speed of sound to be 340 meters per second?
So use the exact sort of analysis that we already described
and see if you can do this.
See if you can find what the lowest frequency is,
the deepest sound that we can find for a given pipe like this one
with two open ends and one meter long.
The way this problem works looks like this.
We are going to first draw our pipe.
So here's our nice pipe.
We are going to again,
sort of use a dotted line just to represent the middle point,
were we have zero pressure.
And again, since this is a pipe with two open ends.
We require at the open ends that there is zero atmosphere pressure.
So we can go ahead and put these two dots here
because we know that whatever wave I draw in here,
it needs to have these two zero points.
So for a wave, what I am looking for is the longest frequency
that I can possibly draw.
And for a wave and just sort of picture this wave to yourself
and ask yourself, what is the longest wavelength I can put in here?
For a given wave like this one, we can see the places that it cross to zero.
So for the longest wavelength,
what I would like is to pick the two closest zeros to each other.
So I can pick this zero and this zero.
In that case, our wave looks like this.
We have no pressure.
We have atmospheric pressure that is at these ends.
Nothing above or below atmosphere pressure.
We have a maximum of pressure right between.
Notice also that we could also have drawn the lower wave.
And these are equivalent because as time goes on,
this point here will go from maximum pressure to minimum pressure
to maximum pressure and back and forth.
So we'll have an oscillation here.
So it doesn't actually matter which one of these you chose.
So now that we've done this, what we'll do is we'll complete the wave
just so we can sort of compare visually
the wavelength with the length of the pipe.
And if you do that and we call the length of this pipe, L.
We can see that this wavelength of the pipe lambda
is going to be twice the length of the pipe itself
and so we have lambda is 2 times L.
So this great, in terms of the length of the pipe,
which is something that we're told in this problem.
We now know the wavelength of our wave
but what we're looking for is the frequency.
So how can we go from the wavelength to the frequency?
We know from our velocity equation that the velocity
is equal to the wavelength of a wave times its frequency
and I'm gonna box this again
because this is a important equation that comes up very, very, very often.
So make sure you are aware of this equation
and know what other variables in it are.
So let's solve that for f the frequency,
we have the velocity of our wave divided by how long the wave is.
We know the speed of sound and air
and we're ready to go ahead and plug in things that we know.
So this is the velocity divided by the wavelength
which we found was twice the length of the pipe.
We can plug in all the numbers that we're given.
This is 340 meters per second divided by twice the length of the pipe
where the length of the pipe was one meter.
So let's just say this here, length is one meter,
make sure we have that written.
So this is divided by 2 meters.
Fortunately, we know that our units cancelled properly.
We get 340 divided by 2 or 170 inverse seconds
1 over second is our unit and again this is not a 5,
be careful here and this is Hertz, so this our frequency, 170 Hz.
So this is how a typical example would work for a pipe.
What you have is your two ends. It'll be opened or closed.
Maybe one open, maybe one closed.
Maybe both open, maybe both closed
and you'll just apply the boundary conditions as we saw.
It might be helpful for you to draw a wave like I drew here,
just you can visualize where are the peaks of a wave,
where are the troughs of a wave, where are the zero points of a wave
and which of these points should I pick
if I want the longest wavelength to fit in my pipe.
Once you've done that,
you'll almost always need to use this velocity equation
depending in what parameter in the question you're trying to find.
If you're just trying to find the wavelength of your wave,
you wouldn't need to use the velocity equation to find the wavelength
as we did here since we needed to find the frequency.
So this is a basic example one more time and again,
this is potentially one of the more confusing topics.
So be sure to go over this a few times.
Try some practice problems
and see if you can make sure that this pipe,
sort of example makes some sort of sense.