Question

A small ball of density ${\rho }_0$ is released from rest from the surface of a liquid whose density varies with depth $h$ as $\rho =\frac{{\rho }_0}{2}(\alpha +\beta h)$. Mass of the ball is $m$. Select the most appropriate option.
(Here $\alpha$ and $\beta$ are positive constants of proper dimensions with $\alpha <2$)

• A. The particle will execute $\text{SHM}$.
• B. The maximum speed of the ball is $\frac{(2-\alpha )\sqrt{g}}{\sqrt{2\beta }}$.
• C. Both $\left(a\right)$ and $\left(b\right)$ are correct.
• D. Both $\left(a\right)$ and $\left(b\right)$ are wrong.

Answer 1

The correct option is C Both $\left(a\right)$ and $\left(b\right)$ are correct.

At equilibrium,
Weight$=F_B$
where $F_B$ is buoyant force offered by liquid acting opposite to the weight of the body.

$\Rightarrow {\rho }_{0}Vg=V\left[\frac{{\rho }_0}{2}\right(\alpha +\beta h_0\left)\right]g$

$\therefore$ Amplitude$A=h_0=\frac{2-\alpha }{\beta }......\left(1\right)$

Let, the position $h_0$ is changed to the new displaced position, $(h_0+x)$. Net change in buoyant force acting on the body

$F_{\text{net}}=F_b-Vg\rho$ $($upwards$)$

$=Vg\left[\frac{{\rho }_0}{2}\{\alpha +\beta (h_0+x\left)\right\}\right]-Vg\rho$

But, $\rho =\frac{{\rho }_0}{2}(\alpha +\beta h_0)$

$F_{\text{net}}=-\left(\frac{Vg{\rho }_0\beta }{2}\right)x......\left(2\right)$

So, $F\propto -x$

Thus we can say that the motion is simple harmonic.

Acceleration of body in SHM, using $\left(2\right)$

$a=\frac{F_{net}}{m}=\frac{F_{net}}{{\rho }_{0}V}=-\left(\frac{\beta g}{2}\right)x$

$\omega =\sqrt{\frac{\beta g}{2}}......\left(3\right)$ $($Comparing with $a=-{\omega }^{2}x)$

Maximum velocity, using $\left(1\right)$ and $\left(2\right)$ $V_{\text{max}}=A\omega =\frac{2-\alpha }{\beta }\sqrt{\frac{\beta g}{2}}=\frac{(2-\alpha )\sqrt{g}}{\sqrt{2\beta }}$

$\begin{array}{c}\text{Why this question ?}\\ \text{This question not only teaches stepwise}\\ \text{method of calculating angular frequency of any SHM, but}\\ \text{also makes use of archimedes principle (Buoyant force).}\end{array}$