Relating Arc Length and Standing Wave Patterns

In summary, the conversation discusses the physics of standing waves on a string and the application of calculus to understand the relationship between the harmonic number and the length of the string. The conversation also explores the concept of wavelength and its relationship to arc length, as well as the role of nodes and antinodes in determining the length of the string. The conclusion is that the length of the string refers to the total length without any wave on it, and the wave length is the distance between two nodes or antinodes.
  • #1
ThomasMagnus
138
0

Homework Statement



I am currently reviewing the physics of 'standing waves on a string'. I know that for the nth harmonic, the length of the 'string' is [itex]\frac{n\lambda}{2}[/itex]. Instead of just memorizing these, I have been trying to apply my knowledge of Calculus to figure out why these numbers are what they are. However, I am having a bit of trouble. Here's what I have tried to do thus far:

The Attempt at a Solution



If you take the graph of Sin(x), the arc length of the curve between 0 and a can be calculated with the following integral:

[itex]\int \sqrt{1+cos(x)^{2}}dx[/itex] (with limits 0 and a). So one wavelength would be equal to a distance of 2[itex]\pi[/itex]. The integral between 0 and [itex]\pi[/itex] [itex]\approx 3.82 [/itex] while the integral between 0 and 2[itex]\pi[/itex] [itex]\approx 7.64 [/itex]

So, that means if we had a wave that was a sine function defined between 0 and [itex]\pi[/itex], the wavelength would be 1/2 that of a sin function between 0 and 2[itex]\pi[/itex] and it would have a length of 1/2 that of the function defined between 0 and 2[itex]\pi[/itex].

What I am confused about is that I keep just getting:

First Harmonic [itex]\frac{\lambda}{2}[/itex]=[itex]\frac{L}{2}[/itex] ([itex]\lambda[/itex]=L)

Apart from it being the wrong answer, this would mean the wavelength, [itex]\pi[/itex], is equal to L which by the arc length formula is 3.82. Furthermore, would the wavelength have to be the same as the arc length (i.e. length of the string?)

Also if I said I wanted to find the length of the sine wave between 0 and [itex]\pi[/itex] wouldn't I be able to say that the wavelength= [itex]\pi[/itex] and by definition L=wavelength/2 for first harmonic. Then I would get [itex]\pi[/itex]/2 = 3.82. 1.57≠3.82

Maybe by Length of the string they don't mean arc length?

Am I just completely missing something here, or am I using Calculus in the wrong place.

Thanks!(EDIT) I think I figure it out. I was making it WAY more complicated then it has to be. Since there are two nodes on each end, it is like fixing a string in two positions and pulling it up, so the length of the string can't change then.
 
Last edited:
Physics news on Phys.org
  • #2
(EDIT) I think I figure it out. I was making it WAY more complicated then it has to be. Since there are two nodes on each end, it is like fixing a string in two positions and pulling it up, so the length of the string can't change then.
Well done :)

Basically the model is for those situations where the overall length of the string can change while the tension remains constant and the separation of the endpoints is a constant ... where there are not a lot of losses. If there are no losses, the standing wave does not even have to be a sine wave. (i.e. pluck it in the middle and you have a triangular standing wave.)

The equation you start with for calculus is the wave equation in one dimension:
http://en.wikipedia.org/wiki/Wave_equation
... the solution is generally not trivial (from scratch) but the upshot is that the solutions can be expanded in a trigonometric series, and you are back to counting nodes and antinodes between 0 and L.

Aren't you glad you asked ?!
 
  • #3
ThomasMagnus said:
Maybe by Length of the string they don't mean arc length?

Correct. The wave length λ is the straight line distance between two nodes (or antinodes). The length of string refers to the length with no wave on it.

Edit: oops my post crossed with Simons.
 

Related to Relating Arc Length and Standing Wave Patterns

1. What is the relationship between arc length and standing wave patterns?

The arc length of a standing wave pattern is directly proportional to the wavelength of the wave. This means that as the arc length increases, the wavelength also increases, and vice versa.

2. How does the arc length affect the frequency of a standing wave?

The frequency of a standing wave is inversely proportional to the arc length. This means that as the arc length increases, the frequency decreases, and vice versa.

3. What is the significance of understanding the relationship between arc length and standing wave patterns?

Understanding this relationship can help scientists and engineers in designing and building structures that can withstand standing wave patterns, such as bridges or buildings. It also helps in the study and analysis of various wave phenomena.

4. Can the arc length of a standing wave be changed?

Yes, the arc length of a standing wave can be changed by altering the length of the medium where the wave is traveling. This can be done by adjusting the tension, density, or length of the medium.

5. How does the number of nodes affect the arc length of a standing wave?

The number of nodes in a standing wave does not directly affect the arc length. However, the number of nodes is related to the wavelength of the wave, which in turn affects the arc length. As the number of nodes increases, the wavelength also increases, resulting in a longer arc length.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
389
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
602
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
937
  • Introductory Physics Homework Help
Replies
2
Views
865
  • Introductory Physics Homework Help
Replies
4
Views
914
  • Introductory Physics Homework Help
Replies
28
Views
389
  • Introductory Physics Homework Help
Replies
3
Views
840
Back
Top