The Oscillation Of Floating Bodies

The Oscillation of Floating Bodies

Consider a solid cylinder of density $\sigma$ and height h, is floating in a liquid of density $\rho$ as shown below figure, And $(\sigma<\rho)$.

If l is the length of the cylinder dipping in liquid as shown in the above figure.

If it is depressed slightly and allowed to oscillate vertically.

Then the time period of the oscillation is given by

$T=2 \pi \sqrt{\frac{l}{g}}$

The time period of the oscillation of the above SHM is also given in terms of $h, \rho, \sigma$

$\begin{aligned} & \text { at mean position } \\ & F_{\text {net }}=0 \Rightarrow \text { Weight of } \text { solid }=\text { buoyant force } \Rightarrow m g=V \rho g \\ & \text { As } m=\sigma h A \\ & \Rightarrow \sigma h A g=\rho l A g \\ & \Rightarrow l=\frac{h \sigma}{\rho}\end{aligned}$

So time period of the oscillation is given by

$T=2 \pi \sqrt{\frac{h \sigma}{g \rho}}$

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Solved Examples Based on the Oscillation of Floating Bodies

Example 1: A rectangular block of mass $m$ and area of cross-section A floats in a liquid of density $\rho$. It is given a small vertical displacement from equilibrium it undergoes oscillation with time period $T$. Then

1) $T \alpha \frac{1}{\rho}$
2) $T \alpha \frac{1}{\sqrt{m}}$
3) $T \alpha \sqrt{\rho}$
4) $T \alpha \frac{1}{\sqrt{A}}$

Solution:

The time period of SHM of small vertical oscillations in a liquid is given by

$
T=2 \pi \sqrt{\frac{l}{g}}
$

where $l$ is the length of the cube/cylinder/block dipped in the liquid.
So according to the law of floatation, the weight of the block $=$ weight of the liquid displaced

$
\begin{aligned}
& \mathrm{mg}=A l \rho g \\
& \Rightarrow l=\frac{\mathrm{m}}{\mathrm{A} \rho} \\
& \Rightarrow \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A} \rho g}} \\
& \Rightarrow T \alpha \frac{1}{\sqrt{A}}
\end{aligned}
$

Hence, the answer is the option (4).

Example 2: Consider a solid cylinder of density $\sigma$ and total height h , is floating in a liquid of density $\rho$ as shown below figure.And $(\sigma<\rho)$.

If l is the length of the cylinder dipping in liquid as shown in the above figure.

Then what is the relation between l and h

1) $l=\frac{h}{2}$
2) $l=\frac{h \sigma}{\rho}$
3) $l=\frac{h \rho}{\sigma}$
4) $h=l$

Solution:

Time Period of the floating body

A floating body is in a stable equilibrium. When it is displaced up and released, it accelerates down and when it is pushed down
and released, it accelerates up. It means a floating body experiences a net force towards its stable equilibrium position. Hence, a floating body oscillates when displaced up or down from its mean position.

Consider a solid cylinder of density $\sigma$ and height h , is floating in a liquid of density $\rho$ as shown below figure, And $(\sigma<\rho)$.

If | is the length of the cylinder dipping in liquid as shown in the above figure.

$
\begin{aligned}
& \text { at mean position } \\
& F_{n e t}=0 \Rightarrow \text { Weight of solid }=\text { buoyant force } \Rightarrow m g=V \rho g \\
& A s m=\sigma h A \\
& \Rightarrow \sigma h A g=\rho l A g \\
& \Rightarrow l=\frac{h \sigma}{\rho}
\end{aligned}
$

Hence, the answer is the option (2).

Example 3: A cylindrical block of wood (density=650 kg m^-3), of base area 30 cm² and height 54 cm, floats in a liquid of density 900 kg m^-3. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly) : (in cm)

1) 39

2) 52

3) 65

4) 26

Solution:

Let Block float with h depth in water.

At equilibrium
$
\left(A H \cdot\left(\rho_B\right) g\right)=A h\left(\rho_l\right) g
$
Let it be depressed by $x$

$
\begin{aligned}
& \Rightarrow f_{\text {net }}=\left(M_{\text {block }} \times g\right)-f_{u p} \\
& f_{\text {net }}=A H\left(\rho_B\right) g-\left(\rho_l\right) \cdot g \cdot A(h+x) \\
& =-A x\left(\rho_l\right) g \\
& \Rightarrow A H \cdot\left(\rho_B\right) \frac{d^2 x}{d t^2}=-A x\left(\rho_l\right) g \\
& \Rightarrow \frac{d^2 x}{d t^2}=-\left(\frac{\left(\rho_l\right) g}{H \cdot\left(\rho_{\text {Block })}\right.}\right) \cdot x \\
& \omega^2=\frac{\left(\rho_l\right) g}{H\left(\rho_B\right)}=\frac{g}{l} \\
& \Rightarrow l=\frac{H\left(\rho_B\right)}{\left(\rho_l\right)}=\frac{650 \times 54}{900}=39 \mathrm{~cm}
\end{aligned}
$

Hence, the answer is the option (1).

Example 4: A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is 2.5 cm, then $\omega$ ( rad s^-1) is close to : (density of water = 10³ Kg/m³)

1) 7.90

2) 2.50

3) 1.25

4) 3.66

Solution:

Equation of S.H.M.
$
\begin{aligned}
a & =-\frac{d^2 x}{d t^2}=-w^2 x \\
w & =\sqrt{\frac{k}{m}}
\end{aligned}
$

wherein

$
x=A \sin (w t+\delta)
$
According to the question,

$
\begin{aligned}
& A \times \rho g=F_{\text {res }} \Rightarrow\left(\pi r^2 \rho g\right) \times=F_{\text {rest }} \\
& \omega^2=\frac{\pi r^2 \rho g}{m} \Rightarrow \omega=r \sqrt{\frac{\pi g}{v}} \\
& \text { since } m=\rho V \\
& \omega=2.5 \times 10^{-2} \sqrt{\frac{3.14 \times 10}{310 \times 10^{-6}}}=2.5 \sqrt{10}=7.90
\end{aligned}
$

Hence, the answer is the option (1).

Summary

In summary, oscillations of floating bodies occur when they are displaced from their equilibrium position, resulting in simple harmonic motion (SHM). The time period of these oscillations depends on the density of the liquid, the dimensions of the floating body, and the depth submerged in the liquid. Various examples, such as floating blocks and cylinders, illustrate how to calculate the time period using SHM principles, highlighting the relationship between force, equilibrium, and oscillatory motion in fluids.