# Building a instrument that plays songs.

1. Apr 8, 2007

### AznBoi

Okay, we need to build an instrument that we can play a song with. The audience needs to recognize the song in order for us to get full points. You can't use any instrument parts and the instrument must be constructed from scratch. I was thinking about making a tubed instrument like this:

Does anyone know what this type of instrument is called?

Now, in order to make this instrument I need to perform some length and mabye some tube diameter calculations right? My teacher gave us all of the note frequencies and we also calculated the speed of sound in the classroom.
Am I suppose to use the equation: $$v=f\lambda$$ to calculate the wave lengths? I know the length formula for tubes with one end closed: $$l = \frac{1}{4}\lambda$$ -- for the 1st harmonic

The thing is I don't know what formulas I need to use to compute the length of the tubes. Do you divide the speed of sound (343m/s) by the note frequencies and then multiply the resulting wavelength in this equation: $$l = \frac{2n-1}{4}\lambda$$?? How do you know what nth term to use? For example, for the note C, it has a frequency of 512 Hz. So can someone please show me an example of how I can calculate the length of a tube with one end closed that will produce that note?? Do I need to take the tube's diameter or anything into mind? I just don't get how fundamental/nth harmonics are related to notes and which ones goes with which. Thanks!

2. Apr 8, 2007

### AznBoi

Does anyone know how notes and harmonics are related??

3. Apr 8, 2007

### gabee

When you play a note on a piano, say the 440 Hz "A" note, that 440 Hz frequency is the fundamental frequency or first harmonic of the note. That means that it is the loudest and most recognizable frequency present in the frequency spectrum of the sound of that note. However, there are also other frequencies present--there is the 880 Hz second harmonic, 1320 Hz third harmonic, 1760 Hz fourth harmonic, etc. present in the full soundwave. The certain combination and amplitude of the various harmonics that make up the soundwave give different instruments their characteristic timbres--it is largely because of these harmonic differences that you can distinguish the sound of a flute from that of a violin or a clarinet.

http://en.wikipedia.org/wiki/Harmonic

Here's an example of a spectrograph analysis for a piano note. The horizontal axis is frequency, the vertical axis is amplitude. Notice how the first harmonic is the loudest and the harmonics occur at regular intervals along the frequency axis.

Last edited: Apr 8, 2007
4. Apr 9, 2007

### AznBoi

I think I'm understanding the harmonics and frequency. For every frequency associate with every note there is a certain wavelength that sound needs to travel in order to produce the maximum note?

However, I still don't know how to calculate the length that the tubes need to be. Am I suppose to use: $$l = \frac{2n-1}{4}\lambda$$ along with $$v=f*\lambda$$ in conjunction to solve for the lengths of the tube? Also, how do you know what nth term to use? How do I calculate the lengths of the tubes for every note (given the frequency of each note and speed of sound)?? Please help, Thanks.

5. Apr 9, 2007

### denverdoc

i think those are pan pipes, south american instrument, very mellow overtones, in other words the even harmonics are predominant.

6. Apr 9, 2007

### AznBoi

So am I suppose to find the fundamental frequency: $$l=1(\frac{1}{4} \lambda)$$ for each note given the frequency of each one?

Solve for $$\lambda$$ you would get:
$$\lambda= 4l$$
Substituting $$\lambda$$ in the equation $$v=f* \lambda$$ you would get: $$v=f*(4l)$$ which is equal to: $$l= \frac{v}{f(4)}$$

So, would the length of each tube for every note depend on the equation:
$$l= \frac{v}{f(4)}$$

$$v$$ is the speed of sound ~343.9 m/s correct? $$f$$ would be the frequency of each note? $$l$$ would be the length of the tube?

I don't get what each of the harmonics are for, can I just use one of the harmonics to get every note. Make the independent variable just the frequency and solve for the length with the equation above? Will that give me every note accurately or what?

Here is the chart of frequencies my teacher gave us:
Note-Frequency(Hz)
C-512Hz
B-480Hz
A-426Hz
G-384Hz
F-341.3 Hz
E-320 Hz
D-288 Hz
C-256 Hz

Last edited: Apr 9, 2007
7. Apr 9, 2007

### AznBoi

Wait, should I use the frequencies on this page instead of the ones my teacher gave me? http://www.phys.unsw.edu.au/jw/notes.html

The frequencies for the notes are all the same right? They don't vary unless you try to find them yourself (like what my teacher did??)? Using the frequencies from the site should give me a more correct-pitched note right?

I know that the speed of sound varies depending on temperature and the surrounding environment but does frequency differ the same way? Are the frequencies given to me by my teacher different from those of the site because he calculated them himself using tuning fork etc??

Every note has a specific frequency right? Does temperature influence frequency? I know that frequency is the # of vibrations or cycles per second and that it can be affected if the wavelengths (and thus the length of the pipe) are affected.

So every specific note frequency codes for a specific length of tube that plays that note?

Should I use the note frequencies that my teacher gave me or the ones that are professionally posted on the site? Which will give me the best correct sounding note?

Last edited: Apr 9, 2007
8. Apr 9, 2007

### AlephZero

There are two things here.

1. The absolute pitch of notes only matters if you want to have several instruments "in tune" with each other. The relative pitch of the notes is more important for getting an instrument that sounds good.

Most modern western musical instruments are built to a standard of A = 440 Hz but that has varied between about A = 400 and A = 455 at different times in history. Your teachers pitches are based on an old (19th century) scientific standard of C = 256. As far as I know that was never actually used for musical instruments.

2. The relative pitches of the notes define the "scale" (the technical term is "temperament") that your instrument will play in. Your teachers pitches are based on an old scale (there are claims it was invented by Pythagoras, but I don't think there is any proof of that) where the ratios of the different notes are all simple fractions:

D/C = 9/8 E/C = 5/4 F/C = 4/3 G/C = 3/2 etc.

Modern instruments are tuned a different way, where the frequency ratio of each semitone is $$2^{1/12} = 1.05946$$. This system has the advantage that you can play in any key.

Just for comparison, the equal tempered pitches based on C=256 compared with your teachers pitches are

C 512.0 512.0
B 480.0 483.3
A 426.0 430.5
G 384.0 383.6
F 341.3 341.7
E 320.0 322.5
D 288.0 287.4
C 256.0 256.0

Unless you are going to be very accurate making measurements etc, the differences are pretty small.

Last edited: Apr 9, 2007