Standing Waves (Instruments) & Interference interpretation?

I thought the question was asking for the closest distances at which the speakers could be placed for the listener to hear no sound. That would be n=0 and n=1. But you're right, the question is saying "no sound" not "no destructive interference". Sorry about that, I think you have it right.In summary, the conversation covers two problems related to standing waves and wave interference. The first problem involves finding the frequency of a violin string when a finger is placed 1/3 of the way down from the neck end. The second problem involves determining the fourth closest distance between two speakers for which the listener does not hear any sound due to destructive interference. Relevant equations and clarifications are provided for
  • #1
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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  • #2
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?
 
  • #3
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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  • #4
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.
 
  • #5
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?
 
  • #6
kuruman said:
Why is that certain?
Because the distance between the speakers cannot be less than zero.
 
  • #7
Ah yes. I read the question too hastily and misunderstood it.
 

Related to Standing Waves (Instruments) & Interference interpretation?

1. What are standing waves in instruments?

Standing waves in instruments are stationary patterns of vibration that occur when a wave reflects back on itself. This results in certain points on the instrument's surface where the amplitude of vibration is maximized or minimized, creating distinct nodes and antinodes.

2. How do standing waves affect the sound produced by an instrument?

Standing waves can affect the sound produced by an instrument by creating areas of reinforcement or cancellation of sound waves. This can result in changes in the overall tone and quality of the sound, as well as the sustain and volume of the instrument.

3. What is the relationship between standing waves and string length in stringed instruments?

The length of a string on a stringed instrument determines the wavelength of the standing wave that is created. The longer the string, the longer the wavelength and the lower the frequency of the standing wave. This is why longer strings produce lower pitched notes, and vice versa for shorter strings.

4. How does interference impact the interpretation of standing waves in instruments?

Interference occurs when two or more waves interact with each other, resulting in either reinforcement or cancellation of the waves. In the context of standing waves in instruments, interference can affect the amplitude and frequency of the standing wave, ultimately impacting the sound produced by the instrument.

5. Can standing waves be controlled or manipulated in instruments?

Yes, standing waves can be controlled and manipulated in instruments through various techniques such as changing the length or tension of strings, altering the shape or material of the instrument, or using dampening materials. This can be done to achieve specific sound characteristics or to correct any unwanted interference effects.

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