Given, Mass of each block A and B, m = 0.1 kg

Radius of circle, $R=0.06 \mathrm{m}$



The natural length of spring ${\mathit{\text{l}}}_0=\text{0.06}\pi =\pi \text{R}$ (Half circle) and spring constant,

$\mathrm{k}=0.1 \mathrm{N} {\mathrm{m}}^{-1}$

In the stretched position elongation in each spring

$\text{x}=\text{R}\theta$

Let us draw FBD of A



Spring in lower side is stretched by 2x and on upper side compressed by 2x. Therefore, a force 2kx on each block would be exerted by each spring.

Hence, a restoring force, F = 4kx will act on A in the direction shown in figure.

Restoring torque of this force about origin

$\tau =-\text{F. R}=-\left(4\text{kx}\right)\text{R}=-\left(4\text{kR}\theta \right)\text{R}$

or $\tau =-4{\text{kR}}^2·\theta$   ..... (i)

Since, $\tau \propto -\theta$, each ball executes angular SHM about origin O.

Equation. (i) can be written as

$\text{I}\alpha =-4{\text{kR}}^2\theta$

or   $\left({\text{mR}}^2\right)\alpha =-4{\text{kR}}^2\theta$

or $\alpha =-\left(\frac{4\text{k}}{\text{m}}\right)\theta$

Frequency of oscillation, $\text{f}=\frac{1}{2\pi }\sqrt{\left|\frac{\text{acceleration}}{\text{displacement}}\right|}$

                                         $=\frac{1}{2\pi }\sqrt{\left|\frac{\alpha }{\theta }\right|}$

                                       $\text{f}=\frac{1}{2\pi }\sqrt{\frac{4\text{k}}{\text{m}}}$

Substituting the values, we have

$\text{f}=\frac{1}{2\pi }\sqrt{\frac{4\times \text{0.1}}{\text{0.1}}}=\frac{1}{\pi }\text{ Hz}$