Let initially the cone be in equilibrium i.e. floating

AP EAPCET 2022 - 4th July Morning Shift Physics - Simple Harmonic Motion Question 46 English Explanation 1

$$\begin{aligned} & \text { From the diagram, it is clear that } \\ & \text { Buoyant force }\left(F_b\right)=\text { weight of the cone. }\left(F_w\right) \\ & \qquad \rho_w \times\left(\frac{1}{3} \pi r^2 x\right) \times g=\rho_c \times\left(\frac{1}{3} \pi R^2 H\right) \times g \end{aligned}$$

$$\Rightarrow \quad 2 r^2 x=R^2 H \quad\left(\because \rho_c=\frac{\rho_w}{2}\right)$$

$$\Rightarrow \quad x=\frac{R^2 H}{2 r^2} \quad \text{... (i)}$$

From the diagram,

$$\begin{array}{ll} & \tan \theta=\frac{r}{x}=\frac{R}{H} \\ \Rightarrow \quad & x=\frac{r H}{R} \quad \text{... (ii)} \end{array}$$

From Eqs. (i) and (ii)

$$\begin{gathered} \frac{R^2 H}{2 r^2}=\frac{r H}{R} \\ \Rightarrow \quad \frac{R^3}{2}=r^3 \Rightarrow r=\left(\frac{1}{2}\right)^{1 / 3} R \end{gathered}$$

Now, the cone. is depressed by small distance $$\delta(<< H)$$ and released it executes simple harmonic motion. In simple harmonic motion, the buoyancy force acts as restoring force which is given as

$$F_b=\text { density } \times \text { volume } \times \text { gravity }$$

AP EAPCET 2022 - 4th July Morning Shift Physics - Simple Harmonic Motion Question 46 English Explanation 2

Since, the displacement is very small, the excess submerged shape of cone. can be considered as a cylinder whose volume is given by $$\pi r^2 \delta$$.

$$\therefore \quad F_b=\rho_w \pi r^2 \delta g=\rho_w \pi\left[\left(\frac{1}{2}\right)^{1 / 3} R\right]^2 \delta g$$

Thus, force is equal to restoring force, hence

$$F_b=k \delta=m \omega^2 \delta$$

$$\begin{array}{ll} & \rho_w \pi\left(\frac{1}{4}\right)^{1 / 3} R^2 \delta g=\rho_c \frac{1}{3} \pi R^2 H \delta \omega^2 \\ & 2 \times 3 \times\left(\frac{1}{4}\right)^{1 / 3} \delta g=H \delta \omega^2 \quad\left(\because \rho_c=\frac{\rho_w}{2}\right) \\ \Rightarrow \quad & \frac{6\left(\frac{1}{4}\right)^{1 / 3} g}{H}=\omega^2 \\ \text { or } & \omega=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \Rightarrow 2 \pi f=\sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}} \\ \Rightarrow \quad & f=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H}\left(\frac{1}{4}\right)^{1 / 3}}=\frac{1}{2 \pi} \sqrt{\frac{6 g}{H} \cdot \frac{1}{4^{1 / 3}}} \end{array}$$

Simple Harmonic Motion

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