Waves on a String linear density?

[just.karl]

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!

 

[Kurdt]

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).

 

[just.karl]

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?

 

[Kurdt]

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 $\mu$ will be have an r squared dependence.

 

[just.karl]

I'm not sure what r squared dependence is and what it's referring to.

 

[Kurdt]

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 $\pi r^2 l$ where $r$ is the radius of the string. Since the linear mass density is the mass divided by the length then $\mu=...$ fill in the rest.

 

[just.karl]

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?

 

[Kurdt]

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.

 

[just.karl]

I realized that right after... sorry

Thanks for all your help. I really appreciate it.