Waves on a String linear density?

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Homework Help Overview

The discussion revolves around calculating the ratio of wave speeds in two steel guitar strings with different diameters and tensions. The participants explore the relationship between the diameter of the strings and their linear density, as well as the implications of these properties on wave speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss how to relate the diameter of the strings to linear density and wave speed. There are attempts to simplify the relationship by considering the strings' lengths and diameters, along with questions about the implications of mass and volume on linear density.

Discussion Status

The discussion is active, with participants providing insights into the mathematical relationships involved. Some guidance has been offered regarding the dependence of linear density on the radius of the strings, and there is acknowledgment of the cancellation of length in the calculations. However, there is no explicit consensus on the best approach yet.

Contextual Notes

Participants are working under the constraints of the problem, specifically focusing on the properties of the strings and their effects on wave speed without providing a complete solution. There is an emphasis on understanding the relationships rather than solving for specific values.

just.karl
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Waves on a String "linear density?"

Two steel guitar strings have the same length. String A has a diameter of .5mm and is under 410N of tension. String B has a diameter of 1.0mm and is under a tension of 820N. Find the ratio of the wave speeds, v_a/v_b, in these two strings.

Linear density u=m/L and v=(F/u)^1/2 to find the wave velocity then I would just divide the two to find the ratio.


What I'm confused about is how to I relate the diameter to the linear density equation or if I do at all.
Please help!
 
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The mass of the steel strings will be their volume multiplied by their density. The mass per unit length is just their total mass divided by their length. The one way we vary linear density in guitar strings is to use different diameter strings rather than the same diameter but different density material (which is more difficult).
 
So since they have the same length, could I then just say that String A is 1/2 and string B is 1 for u? In short?
 
just.karl said:
So since they have the same length, could I then just say that String A is 1/2 and string B is 1 for u? In short?

No, because [itex]\mu[/itex] will be have an r squared dependence.
 
I'm not sure what r squared dependence is and what it's referring to.
 
just.karl said:
I'm not sure what r squared dependence is and what it's referring to.

The mass of the string is the string volume times the density. The volume is [itex]\pi r^2 l[/itex] where [itex]r[/itex] is the radius of the string. Since the linear mass density is the mass divided by the length then [itex]\mu=...[/itex] fill in the rest. :smile:
 
I should have know that... Thanks

But for the length can I give it any value great than zero and still have it work out alright?
 
just.karl said:
I should have know that... Thanks

But for the length can I give it any value great than zero and still have it work out alright?

If you try solving the problem you'll find the lengths cancel.
 
I realized that right after... sorry

Thanks for all your help. I really appreciate it.
 

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