= Solution
The magnitude of the gravitational <force> remains approximately $mg$ during a laboratory swing. Its \b[tangential component changes with position], however. For <simple pendulum> length $\ell$ and <angle> $\theta$ measured from the downward vertical,
$$
\boxed{F_t=-mg\sin\theta,\qquad \ddot\theta=-\frac g\ell\sin\theta.}
$$
This restoring component is zero at the lowest point and largest in magnitude at the turning points. It always accelerates the bob towards the lowest point. The radial gravitational component is $mg\cos\theta$; the <string tension> $S$ must additionally supply the <centripetal acceleration>, so $S-mg\cos\theta=mv^2/\ell$. Thus it is the radial resultant, not gravity alone, that bends the trajectory. The bob has maximum speed at the bottom even though its tangential <acceleration> is then zero.
For a small angular <amplitude>, the <small-angle approximation> gives <simple harmonic motion> with <oscillation period> $T=2\pi\sqrt{\ell/g}$. Measure $\ell$ from pivot to the bob's <centre of mass>, time many complete swings, divide the elapsed time by their number, and calculate
$$
\boxed{g=\frac{4\pi^2\ell}{T^2}.}
$$
Repeat with several lengths and plot $T^2$ against $\ell$; the ideal slope is $4\pi^2/g$. The first-order relative uncertainty obeys $\delta g/g\simeq\delta\ell/\ell-2\delta T/T$, with absolute uncertainty contributions combined according to their dependence. Keep the <amplitude> small: the finite-amplitude period is $T=T_0(1+\theta_0^2/16+\cdots)$, with $\theta_0$ in <radians>. Using the longer finite-amplitude period in the small-amplitude formula would underestimate $g$. Pivot friction, a finite bob and support motion are further experimental corrections.<University Physics Volume 1 by William Moebs and others>[§15.4, simple pendulum and small-angle period]{ref}
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