Given: Radius of the sphere: $R$ Half of the sphere is submerged in the pond.

Solution:

First, identify equilibrium conditions by equating the buoyant force and the sphere's weight. Next, consider the forces when the sphere is displaced: the additional buoyant force acts as the restoring force. Use this to set up the equation of motion for simple harmonic motion (SHM). Compare the resulting equation with the standard SHM form to find the angular frequency. Finally, derive the time period using the relationship between angular frequency and the time period of SHM.

Buoyant Force and Equilibrium:

The buoyant force when the sphere is at equilibrium (half-submerged) is equal to the weight of the sphere.

$$ V_{\text {sub }}=\frac{1}{2} \times \frac{4}{3} \pi R^3=\frac{2}{3} \pi R^3 $$

Weight of the sphere:

$$ W=m g=(\rho \times V) g=\rho \times \frac{4}{3} \pi R^3 \times g $$

At equilibrium, the buoyant force $F_b$ is equal to the weight of the displaced water.

$$ F_b=\rho_w \times V_{\text {sub }} \times g=\rho_w \times \frac{2}{3} \pi R^3 \times g $$

Since the sphere is floating with half of it submerged,

$$ \begin{aligned} F_b & =W \\ \rho_w \times \frac{2}{3} \pi R^3 \times g & =\rho \times \frac{4}{3} \pi R^3 \times g \\ \rho_w \times 2 & =\rho \times 4 \\ \rho_w & =2 \rho \end{aligned} $$

Additional buoyant force:

$\Delta F_b=\rho_w \times A \times x \times g$, where $A=\pi R^2$ (cross-sectional area of the sphere).

Net force acting on the sphere when displaced by $x$ is the restoring force,

$$ F=-\Delta F_b=-\rho_w \times \pi R^2 \times x \times g $$

Using $\rho_w=2 \rho$,

$$ F=-2 \rho \times \pi R^2 \times x \times g $$

Since $F=m a$ and here, $a=y$

$$ m y=-2 \rho \times \pi R^2 \times x \times g $$

Using $m=\rho \times \frac{4}{3} \pi R^3$,

$$ \rho \times \frac{4}{3} \pi R^3 \times y=-2 \rho \times \pi R^2 \times x \times g $$

Simplifying

$$ \begin{aligned} & \frac{4}{3} R y=-2 g x \\ & y=-\frac{3}{2 R} g x \end{aligned} $$

Comparison with SHM Equation:

The standard form of SHM equation is $y=-\omega^2 x$, where $\omega$ is the angular frequency.

$$ \omega^2=\frac{3 g}{2 R} $$

$$ \omega=\sqrt{\frac{3 g}{2 R}} $$

The time period $T$ is given by $T=\frac{2 \pi}{\omega}$.

substituting $\omega$,

$$ T=2 \pi \sqrt{\frac{2 R}{3 g}} $$

Simple Harmonic Motion

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