قالب LaTeX: تقرير مختبر الفيزياء

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\begin{labsection}{Objective}
To determine the local gravitational acceleration $g$ by measuring the period of a simple pendulum at various lengths and comparing the result with the accepted value of $g = \SI{9.81}{m/s^2}$.
\end{labsection}

\begin{labsection}{Theory}
For a simple pendulum with small angular displacement ($\theta < 15°$), the period $T$ is:
\begin{equation}
  T = 2\pi\sqrt{\frac{L}{g}}
\end{equation}
where $L$ is the length of the pendulum and $g$ is the gravitational acceleration. Squaring both sides:
\begin{equation}
  T^2 = \frac{4\pi^2}{g} L
\end{equation}
This is linear in $L$. The slope $m = 4\pi^2/g$, so $g = 4\pi^2/m$.
\end{labsection}

\begin{labsection}{Materials and Methods}
\begin{itemize}
  \item Steel bob ($m = \SI{50.0 \pm 0.1}{g}$)
  \item Nylon string ($L$ varied from \SI{0.40}{m} to \SI{1.20}{m})
  \item Digital stopwatch (resolution: \SI{0.01}{s})
  \item Meter stick (resolution: \SI{1}{mm})
\end{itemize}
For each length, 10 full oscillations were timed and repeated 3 times. The period was computed as $T = t_{10}/10$, with uncertainty from the standard deviation of trials.
\end{labsection}

\begin{labsection}{Data}
\begin{table}[H]
  \centering
  \caption{Pendulum period measurements at various lengths.}
  \label{tab:data}
  \begin{tabular}{S[table-format=1.2] S[table-format=1.3] S[table-format=1.3] S[table-format=1.3]}
    \toprule
    \rowcolor{rowalt}
    {$L$ (\si{m})} & {$T$ (\si{s})} & {$\delta T$ (\si{s})} & {$T^2$ (\si{s^2})} \\
    \midrule
    0.40 & 1.268 & 0.008 & 1.608 \\
    0.60 & 1.553 & 0.010 & 2.412 \\
    0.80 & 1.795 & 0.012 & 3.222 \\
    1.00 & 2.007 & 0.009 & 4.028 \\
    1.20 & 2.198 & 0.011 & 4.831 \\
    \bottomrule
  \end{tabular}
\end{table}
\end{labsection}

\begin{labsection}{Analysis}

\begin{figure}[H]
  \centering
  \begin{tikzpicture}
    \begin{axis}[
      width=0.75\textwidth,
      height=8cm,
      xlabel={Length $L$ (\si{m})},
      ylabel={$T^2$ (\si{s^2})},
      xmin=0.3, xmax=1.3,
      ymin=1, ymax=5.5,
      grid=major,
      legend pos=north west,
    ]
      \addplot[only marks, mark=*, mark size=2.5pt, labblue] coordinates {
        (0.40, 1.608)
        (0.60, 2.412)
        (0.80, 3.222)
        (1.00, 4.028)
        (1.20, 4.831)
      };
      \addplot[red!70!black, thick, domain=0.3:1.3] {4.028*x - 0.003};
      \legend{Data, Linear fit}
    \end{axis}
  \end{tikzpicture}
  \caption{$T^2$ vs.\ $L$ with linear regression. Slope $m = \SI{4.028 \pm 0.045}{s^2/m}$.}
  \label{fig:t2-vs-l}
\end{figure}

From the slope of the linear fit:
\[ g = \frac{4\pi^2}{m} = \frac{4\pi^2}{4.028} = \SI{9.80}{m/s^2} \]

Using standard error propagation on $g = 4\pi^2/m$:
\[ \delta g = g \cdot \frac{\delta m}{m} = 9.80 \times \frac{0.045}{4.028} = \SI{0.11}{m/s^2} \]

\textbf{Final result:} $g = \SI{9.80 \pm 0.11}{m/s^2}$
\end{labsection}

\begin{labsection}{Discussion}
Our measured value of $g = \SI{9.80 \pm 0.11}{m/s^2}$ agrees with the accepted value of \SI{9.81}{m/s^2} within experimental uncertainty. The percent error is:
\[ \%\text{error} = \frac{|9.81 - 9.80|}{9.81} \times 100\% = 0.10\% \]
Sources of systematic error include air resistance (which would increase the measured period and decrease the computed $g$) and the finite size of the bob. The small angle approximation is valid since all swings were initiated at $\theta \approx 10°$.
\end{labsection}

\begin{labsection}{Conclusion}
The gravitational acceleration was determined to be $g = \SI{9.80 \pm 0.11}{m/s^2}$, consistent with the accepted value. The linear relationship between $T^2$ and $L$ confirms the simple pendulum model for small oscillations.
\end{labsection}

\end{document}