91Ó°ÊÓ

A uniform solid sphere of density$\frac{1}{2}$ is floating in water.

TheNewton be second law of motion states that ${\mathbf{m}}_{\mathbf{s}}\stackrel{\mathbf{.}\mathbf{.}}{\mathbf{z}}\mathbf{=}{\mathbf{m}}_{\mathbf{s}}\mathbf{g}\mathbf{-}{\mathbf{ÒÏ}}_{\mathbf{Ó\neg }}\mathbf{aV}\mathbf{\left(}\mathbf{z}\mathbf{\right)}\mathbf{g}$.

A uniform solid sphere of density$\frac{1}{2}$is floating in the water.

Let the radius of the sphere be R.

Newton be second law of motion states thatlocalid="1664323424703" $m_s\stackrel{..}{z}=m_{s}g-{\mathrm{ÒÏ}}_{\mathrm{Ó\neg }}aV\left(z\right)g$

$m_s$is sphere mass.

${\mathrm{ÒÏ}}_{\mathrm{Ó\neg }}$is water density.

$V\left(z\right)$is the volume of the sphere beneath surface.

Slice the sphere into a bunch of disks of thickness dz

Radii be $\sqrt{R^2-z^2}$.

$$\begin{gathered} V=\sqrt{{\left(R^2-z^{\text{'}2}\right)}^2}\mathrm{Ï€}dz\text{'}+\frac{1}{2}\frac{4\mathrm{Ï€}}{3}{R}^3 \\ V=\left(R^{2}z-\frac{z^{\text{'}3}}{3}\right)\mathrm{Ï€}+\frac{2\mathrm{Ï€}}{3}{R}^3 \end{gathered}$$

Since $m_s={\mathrm{ÒÏ}}_s\frac{4\mathrm{Ï€}}{3}{R}^3$

Let${\mathrm{ÒÏ}}_s=\frac{1}{2}$

Substitute values mentioned above in the equation mentioned below.

$$\begin{gathered} m_s\stackrel{..}{z}=m_{s}g-{\mathrm{ÒÏ}}_{\mathrm{Ó\neg }}aV\left(z\right)g \\ \stackrel{..}{z}=\left(1-\frac{V\left(z\right)}{m_s}\right)g \\ \stackrel{..}{z}=\left(\frac{-3z}{2R}+\frac{1}{2}{\left(\frac{z}{R}\right)}^3\right) \\ \stackrel{..}{z}=\frac{-g}{R}\left(\frac{3z}{2R}-\frac{1}{2}{\left(\frac{z}{R}\right)}^3\right) \end{gathered}$$

Substitute values mentioned below in above equation.

$$\begin{gathered} \mathrm{ξ}=\frac{z}{R} \\ \mathrm{τ}=\sqrt{\frac{g}{R}}t \end{gathered}$$

The Equation becomes as follows.

$\mathrm{ξ}\text{'}\text{'}\left(\mathrm{τ}\right)=\left(\frac{-3}{2}\mathrm{ξ}+\frac{1}{2}{\left(\mathrm{ξ}\right)}^3\right)$

Initially the sphere is not moving and is just beneath the water surface, the initial condition for$\mathrm{ξ}\left(\mathrm{τ}\right)$are mentioned below.

$$\begin{gathered} z\left(0\right)=R \\ \stackrel{.}{z}\left(0\right)=0 \\ \mathrm{ξ}\left(0\right)=1 \\ \mathrm{ξ}\text{'}\left(0\right)=0 \end{gathered}$$

Substitute values mentioned above in the equation mentioned below.

The equation becomes as follows

$$\begin{gathered} \frac{1}{2}\mathrm{ξ}{\text{'}}^2-\frac{1}{2}\mathrm{ξ}\text{'}\left(0\right)=\left(\frac{-3}{4}\mathrm{ξ}+\frac{1}{8}{\left(\mathrm{ξ}\right)}^3\right)-\left(\frac{-3}{4}{\mathrm{ξ}}^2\left(0\right)+\frac{1}{8}{\left(\mathrm{ξ}\right)}^4\left(0\right)\right) \\ \mathrm{ξ}{\text{'}}^2=\left(\frac{-3}{2}{\mathrm{ξ}}^2+\frac{1}{4}{\left(\mathrm{ξ}\right)}^4+\frac{5}{4}\right) \\ \mathrm{ξ}{\text{'}}^2=\frac{1}{4}\left(-6{\mathrm{ξ}}^2+{\left(\mathrm{ξ}\right)}^4+5\right) \\ \mathrm{ξ}{\text{'}}^2=\frac{5}{4}(1-{\mathrm{ξ}}^2)(1-\frac{1}{5}{\mathrm{ξ}}^2) \end{gathered}$$

Integrate for period oscillation.

$$\begin{gathered} \underset{1}{\overset{0}{∫}}\frac{d\mathrm{ξ}}{\sqrt{(1-{\mathrm{ξ}}^2)(1-k^2{\mathrm{ξ}}^2})}=-\sqrt{\frac{5}{4}}\underset{0}{\overset{\frac{T}{4}}{∫}}dr \\ T=\frac{8}{\sqrt{5}}K\left(\frac{1}{\sqrt{5}}\right) \\ T≈5.94 \end{gathered}$$

The statement mentioned above states that$\mathrm{ξ}\text{'}\text{'}\left(\mathrm{τ}\right)=\left(\frac{-3}{2}\mathrm{ξ}\right)$.

Substitute the values in the equation mentioned above.

$$\begin{gathered} T_{80}=2\mathrm{π}\sqrt{\frac{2}{3}} \\ {T}_{80}≈5.13 \end{gathered}$$

The ratio of period becomes as follows.

$$\begin{gathered} \frac{T}{T_{80}}≈\frac{5.94}{5.13} \\ \frac{T}{T_{80}}≈1.16 \end{gathered}$$

Hence, the solution is mentioned below.

The period is approximately 1.16 times the period for small oscillations.