W$2.$ Using the data obtained in Step P$5($frequency versus $F),$ calculate the wave speed for each case by multiplying the frequency by $\frac{L}{2}.$ Explain why we use $\frac{L}{2}$ and not $\frac{L}{4}.$ Using Excel, superimpose plots of $v^2$ versus $F$ and $v$ versus $F($you must add a secondary $y-$axis to properly display both sets of data$).$ The horizontal scale and points will be the same in both cases, but the $y$ axes will have two scales $($one for $v$ and another for $v^2).$ Using Excel, add a linear "Trendline" to both sets of data. Can you conclude from looking at this graph whether your data is closer to obeying

$v=aF,or,v=b(F{)}^{\frac{1}{2}}$

where a and $b$ are constants of proportionality. Suppose you thought the theoretical relationship between the tension and the resultant wave speed might indicate an equation of the form $v=c(F{)}^{\frac{1}{3}},$ where $c$ is another constant of proportionality? What quantities would you plot in order to test this $($hint: you would like to find a straight line$)?$ What quantities would you plot to test whether $v=d(F{)}^{-\frac{2}{3}}$ is a valid formula or not, where $d$ is yet another constant of proportionality? Explain your reasoning.

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