91Ó°ÊÓ

We have, Height of the object is h and its massM and cross-sectional area isA

Formula for buoyant force on the body is,

$\mathit{F}\mathbf{=}\mathit{ÒÏ}\mathit{V}\mathit{g}$

Here,

F is buoyant force

P is density of fluid

V is volume of displaced fluid

g is acceleration due to gravity

Formula to calculate force is,

$\mathit{F}\mathbf{=}\mathit{M}\mathit{g}$

Mis mass of object.

Using above formulas we get,

$$\begin{gathered} \mathit{M}\mathit{g}\mathbf{=}\mathit{ÒÏ}\mathit{V}\mathit{g} \\ \mathit{M}\mathbf{=}\mathit{ÒÏ}\mathit{V} \end{gathered}$$

Thus, relation for mass is,

$\mathit{M}\mathbf{=}\mathit{ÒÏ}\mathit{V}$

Formula to calculate volume is,

$\mathit{V}\mathbf{=}\mathit{A}\mathit{x}$

Here,

Vis volume

Ais area

xis length

Again, we get,

$$\begin{gathered} \mathit{M}\mathbf{=}\mathit{ÒÏ}\mathit{A}\mathit{x} \\ \mathit{x}\mathbf{=}\mathit{M}\mathbf{/}\mathit{ÒÏ}\mathit{A} \end{gathered}$$

Therefore, the vertical distance from the surface of the liquid to the bottom of the floated object at equilibrium is $\mathit{M}\mathbf{/}\mathit{ÒÏ}\mathit{A}$.

We have, Height of the object ishand its massM and cross-sectional area isA

The density of liquid is$\mathit{ÒÏ}$

Formula to calculate downward force is,

$\mathit{F}\mathbf{=}\mathit{A}\mathit{x}\mathbf{\text{'}}\mathit{ÒÏ}\mathit{g}$

$\mathit{x}\mathbf{\text{'}}$is farther downward displacement

Rearrange for$\mathit{x}\mathbf{\text{'}}$

$\mathit{x}\mathbf{\text{'}}\mathbf{=}\mathit{F}\mathbf{/}\mathit{A}\mathit{ÒÏ}\mathbf{}\mathit{g}$

Therefore, the position of the bottom of the object in the new equilibrium position is$\mathit{F}\mathbf{/}\mathit{A}\mathit{ÒÏ}\mathbf{}\mathit{g}$

We have, Height of the object ishand its massM and cross-sectional area isA

and the density of the liquid is$\mathit{ÒÏ}$

Formula to calculate spring constant is,

$\mathit{k}\mathbf{=}\frac{\mathbf{F}}{\mathbf{x}}$

x is spring constant

Substitute $\mathit{A}\mathit{x}\mathit{ÒÏ}\mathit{g}$ for F to find k,

$$\begin{gathered} \mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}}{\sqrt{{\displaystyle \frac{\mathbf{A}\mathbf{ÒÏ}\mathbf{g}}{\mathbf{M}}}}} \\ \mathbf{}\mathbf{=}\mathbf{2}\mathbf{Ï€}\sqrt{\frac{\mathbf{M}}{\mathbf{AÒÏg}}} \end{gathered}$$

Formula for angular frequency is,

$\mathit{Ó\neg }\mathbf{=}\sqrt{\mathbf{k}\mathbf{/}\mathbf{M}}$

$\mathit{Ó\neg }$is angular frequency

Substitute$\mathit{A}\mathit{ÒÏ}\mathit{g}$fork to find$\mathit{Ó\neg }$

$\mathit{Ó\neg }\mathbf{=}\sqrt{\frac{\mathbf{A}\mathbf{ÒÏ}\mathbf{g}}{\mathbf{M}}}$

Formula to calculate time period is,

$\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}}{\mathbf{Ó\neg }}$

Substitute$\sqrt{\mathbf{A}\mathbf{ÒÏ}\mathbf{g}\mathbf{/}\mathbf{M}}$for$\mathit{Ó\neg }$to findT

$$\begin{gathered} \mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{Ï€}}{\sqrt{{\displaystyle \frac{\mathbf{A}\mathbf{ÒÏ}\mathbf{g}}{\mathbf{M}}}}} \\ \mathbf{}\mathbf{=}\mathbf{2}\mathbf{Ï€}\sqrt{\frac{\mathbf{M}}{\mathbf{AÒÏg}}} \end{gathered}$$

Therefore, the period of the SHM in terms of the density of liquid and mass and the cross-sectional area of the object is $\mathit{T}\mathbf{=}\mathbf{2}\mathit{Ï€}\sqrt{\mathbf{M}\mathbf{/}\mathbf{A}\mathbf{ÒÏ}\mathbf{g}}$.