How Do You Calculate Fret Placement and Overtones in Musical Instruments?

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To calculate fret placement for a guitar string tuned to E (330 Hz) to play A (440 Hz), the fret must be placed 0.18 m from the end of the string. The wavelength of the 440 Hz wave on the string is 1.1 m. For sound in air at 20 degrees Celsius, the frequency remains 440 Hz with a wavelength of 0.78 m. For a 2.14 m organ pipe, there are 248 overtones if open and 249 if closed, with calculations involving the speed of sound and pipe length. Shortening the string length increases pitch, and using the relationship between frequency and length can assist in solving these problems.
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1) An unfingered guitar string is 0.73 m long and is tuned to play E above middle C (330 Hz). (a) How from from the end of the string must a fret (and your finger) be placed to play A above middle C (440 Hz)? (b) What is the wavelength on the string of this 440-Hz wave? (c) What are the frequency and wavelength of sound in air at 20 degree's C by this fingered string? Answers: a) 0.18 m, b) 1.1 m c) 440 Hz and 0.78 m

I am really stuck. All I can figure is that you incorporate L = v/2f... and f = v/wavelength... help me, please.

2) How many overtones are present within the audible range for a 2.14-m-long organ pipe at 20 degree's if (a) it is open (b) it is closed. Answers: a) 248 overtones, 249 overtones

What I thought you could do was v/2L and v/4L... 343/(2 x 2.14) = 80 and 40... not correct.
 
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See if these examples help - http://www.glenbrook.k12.il.us/gbssci/phys/Class/sound/u11l5b.html

To get a higher pitch, one has to shorten the length of the string.

Try f1 * L1 = f2 * L2, where f is frequency, and then the finger is place x = L1 - L2, and I think you can handle it from here. :smile:
 
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So I would use.

.73 m x 330 Hz = L2 x 440 Hz.

L2 = .547 not the same as the answer 1...
 
Anyone else have a suggestion?
 
Can somone please help me with these problems? :( It is due soon.
 
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