91Ó°ÊÓ

The two spheres are in a simple harmonic motion. For the linear and rotational motions, the force is

$\underset{}{\mathbf{∑}\mathbf{F}\mathbf{}\mathbf{=}\mathbf{-}}\mathit{M}{\mathit{a}}_{\mathbf{c}\mathbf{m}}\mathbf{}\mathbf{=}{\mathit{f}}_{\mathbf{s}\mathbf{-}}{\mathit{k}}_{\mathbf{x}}$

And for the rotational motion, the torque for the system is

$\underset{}{\mathbf{∑}\mathbf{r}\mathbf{=}\mathbf{l}\mathbf{a}}$

Here, is the moment of inertia and for the solid sphere it is given by

$\mathit{I}\mathbf{=}\frac{\mathbf{2}}{\mathbf{5}}\mathit{M}{\mathit{R}}^{\mathbf{2}}$

And$\mathit{Î\pm }$ is the angular acceleration and it is related to the linear acceleration${\mathit{a}}_{\mathbf{c}\mathbf{m}}$by

$\mathit{Î\pm }\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{c}\mathbf{m}}}{\mathbf{R}}$

The torque itself is related to the radius of the share by

$\underset{}{\mathbf{∑}\mathbf{t}\mathbf{a}\mathbf{u}\mathbf{=}\mathbf{R}{\mathbf{f}}_{\mathbf{s}}}$

Herelocalid="1664003486416" ${\mathit{f}}_{\mathbf{s}}$is the friction force.

The relation is given as:

$$\begin{gathered} \underset{}{∑r=}la \\ Rfs=\frac{2}{5}M{R}^2×\frac{a_{cm}}{R} \\ {f}_s=\frac{2}{5}M{a}_{cm} \\ -2m{a}_{cm}=2f_s-kx \\ -2m{a}_{cm}=\left(2×\frac{2}{5}M{a}_{cm}\right)-kx \\ \frac{a_{cm}}{x}=\frac{5}{14}×\frac{k}{M} \end{gathered}$$

The displacement $x\left(t\right)$of a particle from its equilibrium position is described by equation localid="1664009489071" $x\left(t\right)=x\cos \left(\mathrm{Ï–}t+\mathrm{Ï•}\right)$

Here, x is the amplitude of the displacement, localid="1664009470577" $\left(\mathrm{Ï–}t+\mathrm{Ï•}\right)$ is the phase of the motion, and $\mathrm{Ï•}$ is the phase constant.

Take the second derivative you get,

$a=-{\mathrm{Ï–}}^{2}x\cos \left(\mathrm{Ï–}t+\mathrm{Ï•}\right)$

Here the positive term ${\mathrm{Ï–}}^{2}x$ represents the maximum acceleration a when the term $\cos \left(\mathrm{Ï–}t+\mathrm{Ï•}\right)=1$, and you get

localid="1664009953198" $$\begin{gathered} a_{cm}=x{v}^2 \\ =x{\left(\frac{2\mathrm{π}}{t}\right)}^2 \\ \frac{a_{cm}}{x}={\left(\frac{2\mathrm{π}}{t}\right)}^2 \\ {\left(\frac{2\mathrm{π}}{t}\right)}^2=\frac{5}{14}×\frac{k}{M} \end{gathered}$$

$$\begin{gathered} T=2\mathrm{π}\sqrt{\frac{14\mathrm{M}}{5\mathrm{k}}} \\ =2\mathrm{π}\sqrt{\frac{14×0.8}{5×160}} \\ =0.74\mathrm{s} \end{gathered}$$

Therefore, the period in the motion of the centre of mass of the spheres in simple harmonic is 0.74 s.