Standing Waves (Instruments) & Interference interpretation?

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The discussion focuses on two problems related to standing waves and wave interference. For the first question about a violin string, it is confirmed that if a finger is placed 1/3 of the way from the neck, the vibrating length becomes 2/3 of the original, resulting in a new frequency that is 3/2 times the fundamental frequency of 460 Hz. The second question regarding destructive interference from two out-of-phase speakers highlights the ambiguity in the listener's position, as it affects the path difference and resultant sound heard. Clarifications are made regarding the calculation of distances for destructive interference, emphasizing the need to consider the listener's exact location. Overall, the discussion illustrates the complexities of wave behavior in musical instruments and sound interference.
snowcrystal42
Hi,
I'm trying to solve two problems related to standing waves and wave interference; while I'm not having difficulty with the actual solving portion, I don't know if I'm interpreting the questions correctly. Question 1: "A violin string is tuned to 460 Hz (fundamental frequency). When playing the instrument, the violinist puts a finger down on the string 1/3 of the string length from the neck end. What is the frequency of the string when played like this?"

Relevant equations:
v = √(T/μ) where μ is the linear mass density of the string
For a string fixed at both ends L = ½(nλ) or ƒn=(nv)/(2L)

I don't really know much about instruments, but am I correct in thinking that if the string is fingered 1/3 from the neck end, then the vibrating portion will be 2/3 the original length? Which means that the new frequency will be 3/2 as large?

I also have a quick question on wave interference:

Question 2: (A figure is given showing two speakers and a listener located somewhere between them.) "The speakers vibrate out of phase...what is the fourth closest distance to speaker A that speaker B can be located so that the listener hears no sound?"

Relevant equations: For speakers out of phase, destructive interference: ΔL = nλ where n = 0,1,2,3...

Just to check, the "fourth closest distance" includes when n = 0 (when the path differences between the listener and both speakers are the same), right? So I would use n = 3?

Thanks!
 
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snowcrystal42 said:
don't really know much about instruments, but am I correct in thinking that if the string is fingered 1/3 from the neck end, then the vibrating portion will be 2/3 the original length? Which means that the new frequency will be 3/2 as large?
That is correct. The new wavelength is 2/3 of the old wavelength so the new frequency is 3/2 of the old frequency because λf = v = constant. (The speed is constant because the tension is assumed to be the same).
The answer to your second question is "It depends on where the listener is". If the listener is on the perpendicular bisector between the speakers, then the waves will always interfere constructively. So where is the listener? "Somewhere between" is vague.

Edit: Is the listener on the line between speakers?
 
Oops, I should have specified. It looks like this but the numbers are different:
upload_2017-12-12_16-45-27.png
 

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snowcrystal42 said:
the "fourth closest distance" includes when n = 0 (when the path differences between the listener and both speakers are the same), right
Certainly the closest distance is 0, and that does correspond to n=0, but is there another distance for which n=0? Maybe you are not looking for n=3.
 
haruspex said:
Certainly the closest distance is 0 ...
Why is that certain? Assuming that the speakers are driven "out of phase" by π what if distance AC is zero and distance AB is half a wavelength as opposed to a full wavelength?
 
kuruman said:
Why is that certain?
Because the distance between the speakers cannot be less than zero.
 
Ah yes. I read the question too hastily and misunderstood it.
 

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