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So I had a lab yesterday concerning finding a general mathematical expression for the period of compound pendulum. The pendulum let swing freely and time was recorded as well as the length from the point of suspension to the center of mass of the compound pendulum.

When plotting the raw data as it is, a curve like graph was formed. Hence an assumption that the graph $T= 1/x$ was made. However since the graph doesn't quite follow the general trend of the equation given an new assumption is made which is as follow $$T= C_1 l^{-\alpha}+ C_2l^\beta$$ where $C_1, C_2$ are constants. My next mission is to find the exponents so I figured out its just to take the natural logarithm $ln$ of both sides. However that will result in 4 unknowns namely $\alpha, \beta, C_1, C_2$ and The equation I am after is something that looks like $y= mx+c$. I thought that doing demential analysis would solve the issue but unfortunately it didn't:$$T= C_1 l^{-\alpha}+ C_2l^\beta$$ $$s= C_1 m^{-\alpha}+ C_2m^\beta$$ which means C1 and C2 unit has to be $s$ and $\alpha , \beta$ have to cancel each other out. And thats how far I got, I'm lost now. Any suggestion, help, hint will be highly appreciated

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    $\begingroup$ Would you please define the meaning of the quantities $T$, $x$ and $l$? $\endgroup$ – caverac Nov 22 '16 at 9:53
  • $\begingroup$ oh sorry I thought I did, its the Time period and the length from the suspension point to the center of mass point of the compound pendulum $\endgroup$ – Reddevil Nov 22 '16 at 10:04
  • $\begingroup$ I see, in that case you should use $T = C l^{-\alpha}$. That should give you a linear expression once you take $\ln$ in both sides $\endgroup$ – caverac Nov 22 '16 at 10:19
  • $\begingroup$ @caverac if it only was that easy... $\endgroup$ – Reddevil Nov 22 '16 at 12:45
  • $\begingroup$ It is. $\ln T = \ln C - \alpha \ln l = b + m l$, where $b \equiv \ln C$ and $m \equiv -\alpha$ $\endgroup$ – caverac Nov 22 '16 at 12:57
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It is not possible to create a straight-line plot.

The relation between period $T$ and distance $h$ is
$T^2=\frac{4\pi^2}{g}(\frac{k^2}{h}+h)$.
where $I=Mk^2$ is the moment of inertia about the centre of mass. $k$ is called radius of gyration.

This formula cannot be rearranged into a linear plot. Taking logs does not help. If your software is flexible enough you can do a non-linear least-squares curve fit with the function
$y=ax+\frac{b}{x}$
using $y=T^2$ and $x=h$. The software will give you values for parameters $a$ and $b$, from which you can calculate $k$.

Another option (less accurate) is to plot 2 graph of $T^2$ vs $h$ and $T^2$ vs $\frac{1}{h}$ then estimate the slopes for large and small values of $h$ respectively.

Alternatively, you can use the graphical method indicated in the links below. However, as you may appreciate, this is much less accurate. Unlike the least-squares method it relies only on values close to $h=\pm k$, at which the period is a minimum, and makes no use of values further away. The graph is quite flat at these points, which increases uncertainty in the measurement of the distance $2k$ between the minima.

See
School Physics : Compoud Pendulum
Amrita Virtual Lab : Symmetric Compound Pendulum.

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