= Solution
The bob experiences weight $Mg$ vertically downward and a rod <force> towards the pivot. The tangential component of weight is $-Mg\sin\theta$, so
$$
ML\ddot\theta=-Mg\sin\theta\simeq-Mg\theta,\qquad
\boxed{\omega_0^2=g/L,\qquad T_0=2\pi\sqrt{L/g}.}
$$
The rod <force> supplies the radial <acceleration> and contributes no tangential <torque> about the pivot.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-ia2012-pendulum-forces.png]
{title=Original force diagram for a displaced pendulum, with radial and tangential directions}
To the same small-angle order, writing $\chi=\omega_0t+\phi$,
$$
U=MgL(1-\cos\theta)\simeq\frac12MgL\theta_0^2\cos^2\chi,\qquad
K=\frac12ML^2\dot\theta^2=\frac12MgL\theta_0^2\sin^2\chi.
$$
Thus \b[$K+U=MgL\theta_0^2/2$] is constant. Put $b=\dot\theta(0)$, a signed <angular velocity>. Then
$$
\boxed{\theta_0=\sqrt{\alpha^2+b^2/\omega_0^2},\qquad
\phi=\operatorname{atan2}(-b/\omega_0,\alpha)\pmod{2\pi}.}
$$
If the printed <angular speed> $\beta$ is intended as the signed rate, set $b=\beta$. If it means magnitude only, $b=\pm\beta$ gives two possible <phases>; the <amplitude> is unique. At zero <amplitude> the <phase> is arbitrary.
For a <finite-radius spherical pendulum bob>, the PDF dimension $L$ runs from pivot to <centre of mass>. The <parallel axis theorem> gives $I_P=ML^2+2Ma^2/5$. The restoring <torque> remains $-MgL\theta$, hence
$$
T=2\pi\sqrt{\frac{ML^2+2Ma^2/5}{MgL}}
=T_0\sqrt{1+\frac{2a^2}{5L^2}}
\simeq T_0\left(1+\frac{a^2}{5L^2}\right).
$$
With $a=[3M/(4\pi\rho)]^{1/3}=0.02605\,\mathrm m$,
$$
\boxed{100\frac{T-T_0}{T_0}=0.01357\%\simeq0.0136\%.}
$$
The finite bob increases the inertia without changing the gravitational lever arm.
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