A violin string vibrates at 294 Hz when unfingered. At what frequency will it vibrate if it is fingered one third of the way down from the end? f1= 294 Hz (fundamental frequency) f2= 2f1=(2)( 294 Hz)= 588 Hz f3= 3f1=(3)( 294 Hz)= 882 Hz f4= 4f1=(3)( 294 Hz)= 1176 Hz The 294 Hz would be the fundamental frequency and 1/3 the way down would be 1176. Would this be the correct way to solve?
1176? Why do you say that? Think first in terms of wavelength. If one wavelength of 294Hz fits on the full length, what is the ratio of the wavelengths when you press the fret 1/3 of the way down?
I don't think so, but I'm no expert on violins and musical instruments. My interpretation of the question is just to compare the frequency of a fundamental standing wave on a string of length L with the frequency of a fundamental standing wave on a string of length 2L/3. So I draw a horizontal line of length L, and then draw a single sine wave on top of it also of length L. That represents the first situation. Then I draw a mark at 2/3 of L from the left end, and draw a single sine wave of length 2L/3. That represents the shift up in frequency as the string is held down on the fret at the 2L/3 position. What is the ratio of the two wavelengths of the two standing waves? What does that correspond to in terms of the frequency shift? Please show your work.
Third could mean different things. In music theory, in any octave the notes are numbered from 1 to 7. So for example C=1 D=2 E=3 F=4 G=5 A=6 B=7 I don't think this is what you are talking about when you say third octave. I think you are talking about doubling the frequency (octave) and then doubling it again (second octave) and then doubling it yet again (third octave). To do that, you would have to either press the string at a point 7/8 of the way to the bridge, which is probably not even on the fingerboard or alternatively just barely touch the string (not press it) 1/8 of the way from the end to force it to vibrate in a harmonic mode at 8 times the fundamental frequency or 2352 Hz. You often see guitar players using this touch technique to tune guitars. The first (low E) string is touched at 1/4 of its length to produce the second octave that matches the frequency of the sixth (high E) string. Some other touch points are used to match one string frequency to another.
no it would be "the fifth" if that means anything to you. The fifth and the octave are both perfect intervals, in a simple counterpoint composition these intervals are both "correct" starting points.
by "it" you mean the note played when the string is pressed at 1/3 of its length is the fifth (A_4) in the primary or zeroth octave of that string (D_4), where the labels correspond to this table: http://www.phy.mtu.edu/~suits/notefreqs.html I interpreted the third octave to mean the octave three levels above the zeroth octave rather than the third note (F#_4) in the zeroth octave, which would be formed by pressing at 1/5 of the string's length but I'm not absolutely certain of that interpretaton. Pressing at 1/3 the length is not going to give you a third that I can think of.
which part is the end of the violin the shaft or the base... 1/3 towards the bottom you are going to get a rather low Hz
"End" is rather vague, but then how many people know the names of the parts of a violin? How about 1/3 of the way from the nut to the bridge? http://www.phys.unsw.edu.au/~jw/graphics/violin.jpg
If you are at all interested in the physics of music and sound this is one of the best linsk I have ever come across: http://wehner.org/honk/index.htm
The easiest way to see (hear) what is going on when you finger 1/3 the length of a string, is to actually do it. It doesn't need to be a violin to demonstrate the effect. A guitar will do fine. The only difference will be the fundemental frequency of the open string. Here is another reference that may help you understand what is going on.