Help! Guitarist stuck on a math(s) problem!

Question

I'm sure this is embarrassingly basic for the folks around here, but I'm trying to rearrange this equation to give me $uw_2$ - is there anyone who could help? For the record, it's to try and help me figure out the unit weight of a string on a guitar from another string tuned to a different pitch. However, it's the maths bit I'm stuck on! Thanks in advance!

$$ ((uw_1*(2*sl*f_1)^2) / 386.4)/tl_1 - ((uw_1*(2*sl*f_2)^2) / 386.4)/tl_1 = ((uw_2*(2*sl*f_3)^2) / 386.4)/tl_2 - ((uw_2*(2*sl*f_4)^2) / 386.4)/tl_2 $$

Answer 1

You've written

$$ ((uw_1*(2*sl*f_1)^2) / 386.4)/tl_1 - ((uw_1*(2*sl*f_2)^2) / 386.4)/tl_1 = ((uw_2*(2*sl*f_3)^2) / 386.4)/tl_2 - ((uw_2*(2*sl*f_4)^2) / 386.4)/tl_2 $$ which I find hard to read. I'm going to replace "sl" with "S", and $uw_1$, etc., with $u_1$, and $tl_1$ with $T_1$, etc., to get

$$ ((u_1*(2*S*f_1)^2) / 386.4)/T_1 - ((u_1*(2*S*f_2)^2) / 386.4)/T_1 = ((u_2*(2*S*f_3)^2) / 386.4)/T_2 - ((u_2*(2*S*f_4)^2) / 386.4)/T_2 $$

And since you've asked to solve for $u_2$, I'm going to assume $u_1$ is known. First, let's multiply both sides by $T_1 T_2$ and do some cancelling: $$ T_2((u_1*(2*S*f_1)^2) / 386.4) - T_2((u_1*(2*S*f_2)^2) / 386.4)= T_1 ((u_2*(2*S*f_3)^2) / 386.4) - T_1((u_2*(2*S*f_4)^2) / 386.4) $$ Now we can expand $(2*S*f_1)^2$ into $4 S^2 f_1^2$, and let $Q = 4S^2$, because it appears so often. Now we have

$$ T_2((u_1Qf_1^2) / 386.4) - T_2((u_1Qf_2^2) / 386.4)= T_1 ((u_2Qf_3^2) / 386.4) - T_1((u_2*Qf_4^2) / 386.4) $$ Phew! It fits on one line now. Now let's factor out some stuff: $$ (T_2 \cdot u_1 \cdot Q) \left[ \frac{f_1^2}{ 386.4} - \frac{f_2^2}{ 386.4} \right]= (T_1 \cdot u_2 \cdot Q) \left[ \frac{f_3^2}{386.4}- \frac{f_4^2)}{386.4} \right]. $$ Multiply through by $386.4$ to get rid of it everywhere: $$ (T_2 \cdot u_1 \cdot Q) \left[f_1^2 - f_2^2 \right]= (T_1 \cdot u_2 \cdot Q) \left[ f_3^2-f_4^2 \right]. $$ Do a little division: $$ (T_2 \cdot u_1 \cdot Q) \frac{\left[f_1^2 - f_2^2 \right]}{ \left[ f_3^2-f_4^2 \right]}= (T_1 \cdot u_2 \cdot Q). $$ Divide both sides by $Q$: $$ (T_2 \cdot u_1 ) \frac{\left[f_1^2 - f_2^2 \right]}{ \left[ f_3^2-f_4^2 \right]}= (T_1 \cdot u_2 ). $$ and then by $T_1$: $$ (\frac{T_2}{T_1} \cdot u_1 ) \frac{\left[f_1^2 - f_2^2 \right]}{ \left[ f_3^2-f_4^2 \right]}=u_2 . $$

That wasn't so bad, was it?

Answer 2

$$ \frac{uw_1\cdot(2\cdot sl\cdot f_1)^2}{386.4\cdot tl_1 }- \frac{uw_1\cdot (2\cdot sl \cdot f_2)^2)}{386.4\cdot tl_1} = \frac{uw_2 \cdot (2 \cdot sl \cdot f_3)^2}{386.4\cdot tl_2}- \frac{uw_2 \cdot (2 \cdot sl \cdot f_4)^2)}{386.4\cdot tl_2} \implies $$

$$ \frac{uw_1\cdot(2\cdot sl\cdot f_1)^2}{tl_1 }- \frac{uw_1\cdot (2\cdot sl \cdot f_2)^2)}{tl_1} = \frac{uw_2 \cdot (2 \cdot sl \cdot f_3)^2}{tl_2}- \frac{uw_2 \cdot (2 \cdot sl \cdot f_4)^2)}{tl_2} \implies $$

$$ \frac{uw_1\cdot(2\cdot sl\cdot f_1)^2}{tl_1 }- \frac{uw_1\cdot (2\cdot sl \cdot f_2)^2)}{tl_1} =uw_2 \cdot \left[ \frac{(2 \cdot sl \cdot f_3)^2}{tl_2}- \frac{(2 \cdot sl \cdot f_4)^2)}{tl_2}\right] \implies $$

$$ uw_2 = \frac{\frac{uw_1\cdot(2\cdot sl\cdot f_1)^2}{tl_1 }- \frac{uw_1\cdot (2\cdot sl \cdot f_2)^2)}{tl_1}}{\frac{(2 \cdot sl \cdot f_3)^2}{tl_2}- \frac{(2 \cdot sl \cdot f_4)^2)}{tl_2}} \implies $$

$$ uw_2 = uw_1\frac{\frac{f_1^2}{tl_1 }- \frac{f_2^2}{tl_1}}{\frac{f_3^2}{tl_2}- \frac{f_4^2}{tl_2}}=uw_1\frac{tl_2}{tl_1}\frac{f_1^2- f_2^2}{f_3^2- f_4^2}$$

EDIT: Just a side note, in future you should use variable names that are shorter, e.g. only one letter. Instead of using unit weight as $uw_1$ you could have written $w_1$, $tl=l$ and $sl=s$. It makes it easier to see what you have to do.

Answer 3

I hit "enter" to get a new line but that posted the comment. I was then told that I could only edit a comment for 5 seconds! So I decided to post the edited version as an "answer": You can simplify that a lot by factoring out common terms: $\frac{4uw_1sl^2}{386.4tl_1}(f_1^2- f_2^2)= uw_2[\frac{4sl^2}{386.4tl_2}(f_3^2- f_4^2)]$. Now divide both sides by $\frac{4sl^2}{386.4tl_2}(f_3^2- f_4^2)$.

A lot cancels.