Question

A hollow cylindrical shell of radius $R=0.1\text{m}$ has mass $M=6\text{kg}$. It is filled with water having mass $m=3\text{kg}$. It is placed on an inclined plane connected to a spring of spring constant $k=15\text{N/m}$ as shown in the figure. When it is disturbed, it performs oscillations without slipping on the inclined plane. The frequency of the resulting oscillations is
[Assume that the liquid does not rotate inside the cylindrical shell]

• A. $\frac{1}{2\pi }{\text{s}}^{-1}$
• B. $\frac{1}{\pi }{\text{s}}^{-1}$
• C. $\frac{2}{\pi }{\text{s}}^{-1}$
• D. $\frac{1}{3\pi }{\text{s}}^{-1}$

Answer 1

The correct option is A $\frac{1}{2\pi }{\text{s}}^{-1}$


If spring is displaced by $x$ from the equilibrium position, the cylinder will also move by $x$.
Total energy at this instant will be,
$E=P.E\left(\text{cylinder}\right)+P.E\left(\text{spring}\right)+K.E\left(\text{cylinder}\right)$
$\Rightarrow E=-Mg\left(x\sin \theta \right)-mg\left(x\sin \theta \right)+\frac{1}{2}k{x}^2$
$+\frac{1}{2}M{v}^2+\frac{1}{2}M{R}^2\times \frac{v^2}{R^2}+\frac{1}{2}m{v}^2$
$[\omega =vR]$
$\Rightarrow E=-(M+m)gx\sin \theta +\frac{1}{2}k{x}^2+(M+\frac{m}{2})v^2=\text{constant}$
[since the liquid doesn't rotate (given in question), kinetic energy is purely translational]

On differentiating $w.r.t$ time, we get
$\frac{dE}{dt}=-(M+m)g\sin \theta \frac{dx}{dt}+kx\frac{dx}{dt}+2(M+\frac{m}{2})v\frac{dv}{dt}=0$
$\frac{dE}{dt}=0\Rightarrow -(M+m)g\sin \theta +kx+(2M+m)a=0$
$[\because \frac{dx}{dt}=v]$
$\Rightarrow a=-\frac{kx}{(2M+m)}+\frac{(M+m)}{(2M+m)}g\sin \theta$
Let constant $C=\frac{(M+m)}{(2M+m)}g\sin \theta$
So, we have $a=-{\omega }^{2}x+C$
$\therefore \omega =\sqrt{\frac{k}{(2M+m)}}$
So, frequency $f=\frac{1}{2\pi }\sqrt{\frac{k}{(2M+m)}}$
$=\frac{1}{2\pi }\sqrt{\frac{15}{2\times 6+3}}=\frac{1}{2\pi }{\text{s}}^{-1}$

Hence, $\left(a\right)$ is the correct answer.