Wavelength from amplitude information?

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Homework Help Overview

The problem involves a vibrating string at its third-harmonic frequency, where the displacement at a specific point is half the maximum amplitude. The original poster is attempting to determine the length of the string based on this information.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the wave equation and harmonic relationships to find the string length but questions the validity of their results due to an apparent contradiction regarding the string length and displacement point.
  • Some participants suggest potential errors in the calculations, particularly regarding the use of degrees versus radians in trigonometric functions.
  • Others express uncertainty about the correct form of the wave equation to use and the implications of their findings.

Discussion Status

Participants are actively exploring the problem, with some providing guidance on potential calculation errors and the importance of using radians. There is a recognition of conflicting results, and the discussion reflects a mix of interpretations regarding the setup and calculations.

Contextual Notes

There is mention of constraints related to the original poster's limited attempts to answer the question, which may influence their willingness to explore different approaches. The discussion also highlights confusion around the application of trigonometric functions in the context of wave mechanics.

TheExibo
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Homework Statement



A string vibrates at its third-harmonic frequency. The displacement at a point 59.0cm from one end is the first time the displacement is half the maximum amplitude. How long is the string?

Homework Equations



wavelength=(2(rope length))/nodes

y=Asin(kx)

k=2pi/wavelength

The Attempt at a Solution


[/B]
0.5A=Asin(kx)
kx=30

k(0.59m)=30
k=50.85

50.85=2pi/wavelength
wavelength=0.1236m

0.1236m=(2(rope length))/3
rope length=0.1854m

How is such a small rope length possible if 0.59m is half the distance from the first amplitude? What am I doing wrong?
 
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You have noticed that you have a point on the string that is beyond the length of the string ... this is a contradiction: therefore you did something wrong.
This could be arithmetic, maybe you had the calculator in DEG when it should be in RAD or other way?
It could be something to do with the way you reasoned your way through the problem: cannot tell because you don't say... but I did notice that you did your angles in degrees when you should have used radians.
The angles in all scientific formulas are in radians ...

Note: Best practise is to do all the algebra before you plug the numbers in.
 
Last edited:
Simon Bridge said:
You have noticed that you have a point on the string that is beyond the length of the string ... this is a contradiction: therefore you did something wrong.
This could be arithmetic, maybe you had the calculator in DEG when it should be in RAD or other way?
It could be something to do with the way you reasoned your way through the problem: cannot tell because you don't say... but I did notice that you did your angles in degrees when you should have used radians.
The angles in all scientific formulas are in radians ...

Note: Best practise is to do all the algebra before you plug the numbers in.

I used radians, and got a very long length for the wire. I only have one try left for answering my question, so I'm afraid to risk it. My reasoning was to find the wave number using y=Asin(kx) (this equation was suggested in another question, but I don't know if it is right since I've seen y=2Asin(kx) being used), then to use the wave number to find the wavelength, and use the resonance for closed-closed structures to find the length of the rope.
 
I found that y=Asin(kx) should be the correct one to use, and with radians, I get 10.62m. I'm still not sure
 
Never mind, the answer with the radians was correct. How come degrees wouldn't work though?
 

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