W$3($a$)$ Assume that you would like to test Eq$.9\mathrm{.}4$ by using just two data points$-$one with the blue wire and one with the yellow wire. Assume that you know the value of $f$ for the yellow wire with $450$ g of weight suspended from it which produced the $n=5$ mode $($do not use an experimental value for $f$ in this instance; instead, just treat it as an algebraic quantity that is given$).$ If the wave speed is correctly given by $(\frac{F}{}{)}^{\frac{1}{2}}$ and you know the ratio of the densities, $\frac{{}_{\text{b}\text{l}\text{u}\text{e}}}{{}_{\text{y}\text{e}\text{l}\text{l}\text{o}\text{w}}}($use the values given in the table at the end of the Procedure section$),$ then for what total hanging mass would the

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blue string also have five nodes $($i$.$e$.$ four "loops" or anti$-$nodes$)$ at the same frequency that the yellow wire achieved five nodes with a total suspended mass of $450$ g $?$ Show all of your calculations. Hint: write down two versions of Eq$.9\mathrm{.}4,$ one for the blue wire and another for the yellow wire, keeping all of the quantities as algebraic variables, labeled with the subscripts "blue" and "yellow"; since the two frequencies are the same in this case, the two equations can be set equal to each other, and the rest is simply a matter of some algebraic manipulations.

$($b$)$ Calculate the mass that would be required if you were to use a red$-$colored wire with $=11\mathrm{.}200\frac{g}{m}$ instead of the blue$-$colored wire $($see the table of $$ values at the end of the Procedure section$).$

Note: These are theoretical calculations using Eq$.9\mathrm{.}4$ and the values cited in Parts $($a$)$ and $($b$).$ Do not utilize any of the experimental values obtained for these modes, otherwise you will not obtain the correct answer $($i$.$e$.$ due to experimental error, the measured frequencies do not coincide with the theoretical values$).$

P$3.$ Wire Number $1($The Blue Wire$)$

Total mass $($attached plus hanger$)=500g$

$=7\mathrm{.}440glm,L=57\mathrm{.}25cm$

average $v_{\text{p}\text{a}\text{t}\text{e}\text{m}}=13\mathrm{.}614\frac{m}{sec}$

$v_{\text{c}\text{a}\text{l}\text{c}\text{u}\text{l}\text{a}\text{t}\text{e}\text{d}}=\sqrt[\phantom{2}]{m\frac{g}{}}=25\mathrm{.}66324514$

P$4.$ Wire Number $2($The Yellow Wire$)$ Total mass $($attached plus hanger$)=500g$

$=3\mathrm{.}080\frac{g}{m},L=145\mathrm{.}5cm$

average $v_{\text{p}\text{a}\text{t}\text{t}\text{e}\text{r}\text{n}}=20\mathrm{.}5353\frac{m}{sec}$

$v_{\text{c}\text{a}\text{l}\text{c}\text{u}\text{l}\text{a}\text{t}\text{e}\text{d}}=\sqrt[\phantom{2}]{m\frac{g}{}}=39\mathrm{.}8862017S$

P$5.$ Wire Number $2($Yellow Wire$)$ with Varying Hanging Mass