Change guitar string tension given two frequencies

[Sethius]

Homework Statement

A particular guitar string is supposed to vibrate at 217 Hz, but it is measured to actually vibrate at 222 Hz. By what percentage should the tension in the string be changed to get the frequency to the correct value? Do not enter units.

Homework Equations

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}
L=string length
T=Tension
\mu=Linear Density

The Attempt at a Solution

I know that frequency is proportional to the square root of tension given that all other factors remain constant. I solved two frequency equations for \frac{1}{2L} and set them equal to each other with my T_{2}being multiplied by a variable "k" being the conversion factor. my equation looked like this:

\frac{f_{1}}{\sqrt{T_{1}}}=\frac{f_{2}}{\sqrt{kT_{2}}}

I then canceled the \sqrt{T}'s and ended up with \frac{f_{1}}{f_{2}}=\frac{1}{\sqrt{k}}. Solving for k I get 1.0466 which would equal a 4.66% increase in tension however this answer isn't correct. Thank you for any help

 

[BruceW]

Your answer looks good to me... I've asked the other homework helpers to see if they can figure out why your answer isn't correct.

 

[technician]

I got f1/f2 = 0.9775 which gives squrt(k) = 1.0230 which gives k = 1.0114
Is this closer to the answer?

 

[Doc Al]

Solving for k I get 1.0466 which would equal a 4.66% increase in tension however this answer isn't correct.

So you are saying that you must increase the tension to reduce the frequency? Check that. (You just have things reversed a bit.)

 

[Sethius]

Your answer looks good to me... I've asked the other homework helpers to see if they can figure out why your answer isn't correct.

Ok thank you

I got f1/f2 = 0.9775 which gives squrt(k) = 1.0230 which gives k = 1.0114
Is this closer to the answer?

I don't actually have an answer to look at, I have 10 attempts on a LonCapa website to get the right answer. As for your answer, if \sqrt{k}=1.0230 shouldn't you square the 1.0230 as opposed to square rooting it like you've done?

 

[technician]

Sorry! You are correct Seth

 

[Sethius]

So you are saying that you must increase the tension to reduce the frequency? Check that. (You just have things reversed a bit.)

I'm having a hard time wrapping my brain around this. I would assume that the percent change would remain the same regardless of increase or decrease in tension. Should I add a negative sign?

 

[Doc Al]

I would assume that the percent change would remain the same regardless of increase or decrease in tension.

Rethink that. To go from 100 to 110 is a ten percent increase. But to go from 110 to 100 is what percent decrease?

 

[Sethius]

Ok I understand now. So if I'm going from 222Hz to 217Hz I would get .977 as my ratio, then squaring that I would get .955. It's here I'm not entirely sure what to do. Do I subtract that from 1? Or divide one by it?

 

[Doc Al]

Ok I understand now. So if I'm going from 222Hz to 217Hz I would get .977 as my ratio, then squaring that I would get .955. It's here I'm not entirely sure what to do. Do I subtract that from 1? Or divide one by it?

Compared to 1, that's a percent drop of what?

 

[Sethius]

Compared to 1, that's a percent drop of what?

It would be 4.5%?

 

[Doc Al]

It would be 4.5%?

Sounds good. You might what to take that to three sig figs, in case your online system is picky.

 

[Sethius]

Thank you very much for the help!