In the analysis of SHM of a point sized bob oscillating with small angular displacement we can analyse the SHM in both linear and angular terms and arrive at the same answer and this should be true for other rigid bodies executing angular SHM since in all cases $\theta \approx x_{cm}/d$, where d is the distance of COM from point of suspension so if $\theta$ varies as a SHM so is $x_{cm}$ going to.

I tried the same for a solid sphere of radius $R$ and mass $m$ connected with a massless string of length $l$. On displacing the bob by an angle $\theta$ the restoring force acting on the COM is going to be:

$F_r = mgsin(\theta) $

Also, $sin(\theta)\approx \theta = x/l $ where $x$ is the linear displacement of the COM . So: $F_r = mgx/l $

So, the Time period is going to be: $T = 2\pi\sqrt{l/g}$

So I end up getting the same equation as that of the point sized bob..which I guess is wrong because when we analyse this scenario using angular SHM, as I was taught, we get a different answer:

So the torque about the point of suspension P is going to be: $$\tau_p = mgsin(\theta)\times l \approx mgl\theta= I_p \alpha$$ So using $T = 2\pi\sqrt{I_p/C}$, where $C = mgl $

So, the time period here is going to be: $$ T= 2\pi\sqrt{\frac{2mR^2/5 + m(R+l)^2}{mgl}}$$ which of course is different from my previous analysis. So I am clearly missing something ,help will be really appreciated.