Two uniform solid spheres, each with mass$\mathbf{M}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{}\mathbf{kg}$and radius$\mathbf{R}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{0800}\mathbf{}\mathbf{m}$, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant$\mathbf{k}\mathbf{=}\mathbf{160}\mathbf{}\mathbf{N}\mathbf{/}\mathbf{m}$has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the centre of mass of the spheres, which is midway between the centres of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the centre of mass of the spheres is simple harmonic and calculate the period.

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

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

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

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

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

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

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

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

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

The relation is given as:

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

The displacement $x\left(t\right)$of a particle from its equilibrium position is described by equation localid="1664009489071" $x\left(t\right)=x\cos \left(\varpi t+\varphi \right)$

Here, x is the amplitude of the displacement, localid="1664009470577" $\left(\varpi t+\varphi \right)$ is the phase of the motion, and $\varphi$ is the phase constant.

Take the second derivative you get,

$a=-{\varpi }^{2}x\cos \left(\varpi t+\varphi \right)$

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

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

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

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