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?
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?
Ok thank you
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?
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...
Just created an account to answer this. I'm working on the same question now, you should have also been given an equation of the form y(x,t)=Acos(kx-ωt+∅). The co-efficient of t is your omega and the co-efficient of cos is your amplitude. Hope this helped.