Question

A cylindrical block of wood of mass m, radius r and density p is floating in water with its axis vertical. It is depressed a little and then released. If the motion of the block is simple harmonic. Find its frequency.

& A.\,\dfrac{1}{2\pi }\sqrt{\dfrac{\pi {{r}^{2}}pg}{m}} \\\ & B.\,\dfrac{1}{2\pi }\sqrt{\dfrac{\pi {{r}^{3}}pg}{m}} \\\ & C.\,\dfrac{1}{4\pi }\sqrt{\dfrac{\pi {{r}^{2}}pg}{m}} \\\ & D.\,\dfrac{1}{4\pi }\sqrt{\dfrac{\pi {{r}^{3}}pg}{m}} \\\ \end{aligned}$$

Answer 1

The product of the volume of water, the density of water and the acceleration due to gravity equals the weight of the block. The net force on the block at some displacement equals the spring force. Using these equations, we will find the expression for the frequency of the simple harmonic motion.

Complete step by step answer:
From the given information, we have the data as follows.
A cylindrical block of wood of mass m, radius r and density p is floating in water with its axis vertical. It is depressed a little and then released.
Suppose a height h of the block is dipped in the water in an equilibrium position. If ‘r’ be the radius of the cylindrical block, the volume of the water displaced will be $\pi {{r}^{2}}h$. For floating in equilibrium,
$\pi {{r}^{2}}hpg=W$….. (1)
Where p is the density of water and W is the weight of the block.
Now suppose during the vertical motion, the block is further dipped through a distance x at some instant. The volume of the displacement waster is $\pi {{r}^{2}}(h+x)$. The forces acting on the block are the weight W vertically downward and the buoyancy $\pi {{r}^{2}}(h+x)pg$ vertically upward.
The net force on the block at displacement x from the equilibrium position is,

& F=W-\pi {{r}^{2}}(h+x)pg \\\ & \therefore F=W-\pi {{r}^{2}}hpg-\pi {{r}^{2}}xpg \\\ \end{aligned}$$ Using the equation (1) in the above equation, we get, $$\begin{aligned} & F=\pi {{r}^{2}}hpg-\pi {{r}^{2}}hpg-\pi {{r}^{2}}xpg \\\ & \Rightarrow F=-\pi {{r}^{2}}xpg \\\ & \therefore F=-kx\,\,\,\,\,\,(\because k=\pi {{r}^{2}}pg) \\\ \end{aligned}$$ Thus, the block executes SHM with a frequency of, $$v=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}$$ Substitute the value of k in the above equation. $$v=\dfrac{1}{2\pi }\sqrt{\dfrac{\pi {{r}^{2}}pg}{m}}$$ $$\therefore $$ The frequency of the simple harmonic motion of the block is $$\dfrac{1}{2\pi }\sqrt{\dfrac{\pi {{r}^{2}}pg}{m}}$$ **So, the correct answer is “Option A”.** **Note:** We have proved that the block executes simple harmonic motion by expressing the net force on the block at some displacement from the equilibrium position. The product of the volume of water, density of water and the acceleration due to gravity equals the weight of the block.