91影视

We have, mass of the object ismand length of the stick isL.

Formula for frequency of oscillations of the system is,

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{蟺}}\sqrt{\frac{\mathbf{m}\mathbf{g}\mathbf{D}}{\mathbf{l}}}$

mis the mass of square object

gis the acceleration due to gravity

Dis the distance from the hook

Iis the moment of inertia

fis frequency of oscillations

The centre of mass of object is at its geometrical centre. Thus, distance from hook is,

$\mathit{L}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{4}\mathbf{}\mathbf{5}\mathbf{=}\mathit{L}\mathbf{/}\sqrt{\mathbf{2}}$

Lis the length of stick.

From the parallel axis theorem moment of inertia of each stick is,

$$\begin{gathered} \mathit{I}\mathbf{=}\frac{\mathbf{1}}{\mathbf{12}}\mathit{m}{\mathit{L}}^{\mathbf{2}}\mathbf{+}\mathit{m}{\left(\frac{L}{2}\right)}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathit{m}{\mathit{L}}^{\mathbf{2}}\left(\frac{12+4}{12脳4}\right) \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathit{m}{\mathit{L}}^{\mathbf{2}} \end{gathered}$$

Therefore, The moment of inertia of each stick $\mathbf{1}\mathbf{/}\mathbf{3}\mathit{m}{\mathit{L}}^{\mathbf{2}}$.

We have, mass of the object ismand length of the stick isL.

The total moment of inertia for four object is,

$\mathit{l}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{m}{\mathit{L}}^{\mathbf{2}}$

For the entire object and axis at hook, the moment of inertia is found by parallel axis theorem for the objects of mass4m.

$$\begin{gathered} \mathit{I}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{m}{\mathit{L}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathit{m}{\left(\frac{L}{\sqrt{2}}\right)}^{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{10}}{\mathbf{3}}\mathit{m}{\mathit{L}}^{\mathbf{2}} \end{gathered}$$

Substitute$\frac{\mathbf{10}}{\mathbf{3}}\mathit{m}{\mathit{L}}^{\mathbf{2}}$for l

$\mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{蟺}}\sqrt{\frac{\mathbf{m}\mathbf{g}\mathbf{D}}{\left(\frac{10}{3}m{L}^2\right)}}$

Substitute$\frac{\mathbf{L}}{\sqrt{\mathbf{2}}}$forD,

$$\begin{gathered} \mathit{f}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathbf{蟺}}\sqrt{\frac{\mathbf{m}\mathbf{g}\frac{\mathbf{L}}{\sqrt{\mathbf{2}}}}{\left(\frac{10}{3}m{L}^2\right)}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.921}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathit{蟺}\sqrt{\mathbf{g}\mathbf{/}\mathbf{L}\mathbf{)}} \end{gathered}$$

Therefore, frequency at which the object will swing back and forth is $\mathbf{=}\mathbf{0}\mathbf{.921}\mathbf{(}\mathbf{1}\mathbf{/}\mathbf{2}\mathit{蟺}\sqrt{\mathbf{g}\mathbf{/}\mathbf{L}\mathbf{)}}$