$\Delta x=2 \lambda=S_{1} P-S_{2} P$
$\Longrightarrow 2 \lambda=S_{1} S_{2} \cos \theta$
Given $S_{1} S_{2}=3 \lambda$ and $\cos \theta=\frac{D}{\sqrt{x^{2}+D^{2}}}$
Hence $3 \lambda \frac{D}{\sqrt{x^{2}+D^{2}}}=2 \lambda$
$\Longrightarrow 9 D^{2}=4\left(x^{2}+D^{2}\right)$
$\Longrightarrow x=\frac{\sqrt{5} D}{2}$
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$(i)$ acceleration of the centre of mass of ring is $\frac{g}{3}$
$(ii)$ acceleration of the hanging particle is $\frac{2g}{3}$
$(iii)$ frictional force (on the ring) acts along forward direction
$(iv)$ frictional force (on the ring) acts along backward direction
