Length of string on instrument to produce specific frequency

[idkididk]

Homework Statement

You are designing a two-string instrument with metal string 35.0 long, as shown in the figure . Both strings are under the same tension. String has a mass of 8.45 and produces the note middle C (frequency 262 ) in its fundamental mode.

To extend the range of your instrument, you include a fret located just under the strings, but not normally touching them. How far from the upper end should you put this fret so that when you press tightly against it, this string will produce C# (frequency 277 ) in its fundamental?

Homework Equations

f= (1/(2L)) * (F/u)^.5, where f is the frequency, L is length of the string, u is the mass per unit distance, and F is the tension force.

The Attempt at a Solution

So I found that the tension of the string to be 812N. Using that, I solved for x: 277= (1/(2x)) *(812/(.00845/x))^.5

x= .313, which means 31.3 cm. It's asking how far form the upper end, so 35-31.3cm = 3.7cm.

However, this is not correct. Anyone know why?

 

[Doc Al]

So I found that the tension of the string to be 812N.

How did you solve for that? What units are the length and mass given in?

Edit: I see. Length in cm and mass in grams.

 

[Doc Al]

So I found that the tension of the string to be 812N. Using that, I solved for x: 277= (1/(2x)) *(812/(.00845/x))^.5

8.45 is the mass of the full string, so don't divide by the new (smaller) length to find μ. (μ is a constant for the string material, regardless of length.)

 

[idkididk]

^I just tried that, 277= (1/(2x)) *(812/(.00845/.35))^.5, and got x=33cm, so the answer would be 2, but that's not it either

EDIT: Ok nevermind the answer is 1.9 lol. Thanks!