Variable density - SHM

Consider a cuboidal element of fluid of height $\delta y$ and cross section area $\delta A$ at a height $y$ above the bottom of the container. By drawing a free body diagram, one can balance all forces along the vertical direction. Let the pressure on the lower face of this element be $P$ and that on the upper surface be $P + \frac{dP}{dy}\delta y$ .

The force balance looks as such:

$P\delta A = \left(P + \frac{dP}{dy}\delta y\right) \delta A + \rho \delta y \delta Ag$ $\implies \frac{dP}{dy} = -\rho g$

Now, consider the mass to be at a distance $y$ above the bottom of the container. Then, by applying Newton's second law to this small mass gives the following. Consider the mass to have a cross section area of $\delta A_m$ and length of $\delta y_m$ .

$P(\delta A_m) - \left(P + \frac{dP}{dy}(\delta y_m)\right)(\delta A_m) -mg = m\ddot{y}$ $\implies -\frac{dP}{dy}(\delta y_m) (\delta A_m) - mg = m\ddot{y}$

$m = \frac{5}{2}\rho_o (\delta y_m) (\delta A_m)$

Substituting all expressions and simplifying gives:

$\left(4 - \frac{3y}{h_o}\right)g - \frac{5}{2}g = \frac{5}{2}\ddot{y}$

This equation represents an oscillatory motion of frequency:

$\omega^2 = \frac{6g}{5h_o}$

The equilibrium is established when

$\begin{aligned} 2.5 \rho_0 & = \rho_0 \left(4 - \frac {3h}{h_0}\right) \\ \frac 52 & = 4 - \frac {3h}{h_0} \\ \implies h & = \dfrac {h_0}2 \end{aligned}$

When the block overshot $h = \dfrac {h_0}2$ by $x$ , the density of the liquid the block displaced is

$\begin{aligned} \rho & = \rho_0 \left(4 - \frac {3\left(x+ \frac {h_0}2\right)}{h_0}\right) = 2.5\rho_0 - \frac {3\rho_0}{h_0}x \end{aligned}$

This means that there is an restoring force of $\dfrac {3\rho_0Vg}{h_0}x$ , where $V$ is the volume of the block. This must be equal to the $m\ddot x$ , where $\ddot x$ is acceleration of the block. Then we have:

$\begin{aligned} \blue m\ddot x + \frac {3\rho_0Vg}{h_0}x & = 0 & \small \blue{\text{Note that }m = 2.5\rho_0 V} \\ \blue{2.5 \rho_0V} \ddot x + \frac {3\rho_0Vg}{h_0}x & = 0 \\ \ddot{x} + \frac {6g}{5h_0}x & = 0 \end{aligned}$

The frequency of oscillation is then $f = \dfrac 1{2\pi} \sqrt{\dfrac {6g}{5h_0}}$ . Therefore $\alpha = \dfrac 1{2\pi} \sqrt{\dfrac 65} \approx \boxed{0.174}$ .