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Ning Zhu edited section_Analysis_1_Direct_Fit__.tex
over 8 years ago
Commit id: 33630741a2afec8945b16980d8d418c52d918940
deletions | additions
diff --git a/section_Analysis_1_Direct_Fit__.tex b/section_Analysis_1_Direct_Fit__.tex
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$$V_{c}=\frac{1}{L}\times \frac{\Delta\theta}{\Delta B}=\frac{1}{0.1m}\times 2.07\frac{radians}{T}=20.7\frac{radians}{T*m}$$
2. Slope Fit\newline
Initially, we fit our data to the form $V=V_{0}(sin(\theta_{1}+\theta_{2}))^2$, which using trigonometry, is $V=V_{0}\frac{(1-cos(\theta_{1}+\theta_{2})^2)}{2}$ (the actual equation that we fit to). However, we should have been fitting to the form $V=V_{0}cos(\theta_{1}-\theta_{2})^2$, which is $V=V_{0}\frac{(1+cos(\theta_{1}-\theta_{2})^2)}{2}$. Now,
$$sin(\theta_{1}+\theta_{2})^2=sin2(θ1-(-θ2))$$
2π-θ2=θ2’ to account for the incorrect sign in our original equation.
θ2’-π=θ0 to change from sine to cosine, so we do end up with the form
V=V0(½)(1+cos(2(θ1-θ0)))
which is V=V0cos2(θ1-θ2), giving us the form we should have initially fit to.
As shown in Figure 2:
$$V=0.000443B+0.127$$
Therefore: