Waves and Sound - Air Column (Clarinet)

[harujina]

Homework Statement

A clarinet behaves as an air column that is open at one end. For a particular fingering, the length of this air column is 24.6 cm. At 20°C, this fingering sounds the pitch "F" which is 349.2 Hz. During a concert, the breath of the musician raises the bore temperature to 27° C. A single percentage change in frequency would be noticeable to any listener. What would this new frequency be and would it be noticeable?

Homework Equations

v = fλ
v (speed of sound) = 331.4 m/s + (0.606 m/s/°C) T

The Attempt at a Solution

I did this to find speed of sound in 20°C: v (speed of sound) = 331.4 m/s + (0.606 m/s/°C) (20°C)
and I got 343.52 Hz, but it's 349.2 Hz in the clarinet (which is open at one end).
I don't understand this concept; why is it 5.7 Hz higher?
Also, in order to calculate the wavelength, I converted 24.6 cm into .246 m but I forgot what to do with this in the case that it is in a medium with an open end. I remember it's like λ/2 or something?

 

[Simon Bridge]

The speed of sound in air at 20degC is 343.52m/s ... the frequency of the sound was 349.2Hz - therefore, what was the wavelength?

 

[harujina]

The speed of sound in air at 20degC is 343.52m/s ... the frequency of the sound was 349.2Hz - therefore, what was the wavelength?

Ohhh, right...
v = fλ
λ = v/f; = 343.52m/s / 349.2Hz; = 0.98 m?
I thought there was a certain adjustment I had to do for different mediums though? In this case, a clarinet which has one open end.

 

[Simon Bridge]

The medium in question is the air - you do have to adjust the equation for different gas mixtures, but that's unlikely to be an issue here.

The open end, and length, of the clarinet, determines the available wavelengths (the normal modes) that can be excited. The fingering selects from this set of wavelengths by forcing an antinode at particular positions (iirc).

Since you know the length of the air column, and the main wavelength, you can work out which mode is being played.
But you are already told which wavelength has been excited.

 

[harujina]

The medium in question is the air - you do have to adjust the equation for different gas mixtures, but that's unlikely to be an issue here.

The open end, and length, of the clarinet, determines the available wavelengths (the normal modes) that can be excited. The fingering selects from this set of wavelengths by forcing an antinode at particular positions (iirc).

Since you know the length of the air column, and the main wavelength, you can work out which mode is being played.
But you are already told which wavelength has been excited.

I'm confused; how could I know how many antinodes there are/what mode it's playing by the length of the clarinet and the wavelength?
Can't it only be either fundamental or third harmonic?

EDIT: Okay I just tried all the possible medium lengths to produce a wave in a media with a fixed end and open end which are λ/4, 3λ/4, and 5λ/4.
So it's λ/4 (fundamental/first harmonic) because 0.98m/4 = 0.245m, which is the length of the air column.
Is this the only way to check though?

And now I know that n=1 (first harmonic) so I used the equation: f = v/4l (length) and found the frequency to be 336.79 Hz at 27° C (with the speed of sound at 347.762 m/s). I hope this is right...?

 

[Simon Bridge]

Is this the only way to check though?

Since you know $L$ and $\lambda$ and you know the formula for the wavelength of the nth mode, you can just solve for n.

I was expecting that the fingering fixes the wavelength - the speed of sound changes from u to v ...

$u=f_1\lambda$ and $v=f_2\lambda$

You can get both u and v off the temperature formula.

Start with the end in mind:
You want to know the percentage change in frequency ... that would be: $$100 \frac{f_2-f_1}{f_1}$$
If you do all the algebra first, you have less work to do.
If that number is bigger than 1, then you will notice the difference.

The problem statement seems to have a lot of redundant information.

 

[harujina]

Since you know $L$ and $\lambda$ and you know the formula for the wavelength of the nth mode, you can just solve for n.

I was expecting that the fingering fixes the wavelength - the speed of sound changes from u to v ...

$u=f_1\lambda$ and $v=f_2\lambda$

You can get both u and v off the temperature formula.

Start with the end in mind:
You want to know the percentage change in frequency ... that would be: $$100 \frac{f_2-f_1}{f_1}$$
If you do all the algebra first, you have less work to do.
If that number is bigger than 1, then you will notice the difference.

The problem statement seems to have a lot of redundant information.

$$100 \frac{f_2-f_1}{f_2}$$ I thought it was this?
And okay, thank you so much!

 

[Simon Bridge]

Well when you have to do a percentage, you have to ask "percentage of what?"
You are asked to find a percentage change, what is the frequency changing from?