Question

A box $B$ of mass $M$ hangs from an ideal spring of force constant $k$. A small particle $A$ also of mass $M$ is stuck to the ceiling of the box and the system is in equilibrium. The particle gets detached from the ceiling and falls to strike the floor of the box. It takes time $t$ for the particle to hit the floor after it gets detached from the ceiling. The particle on hitting the floor sticks to it and the system thereafter oscillates with a time period of $T$. Find the height $H$ of the box if it is given that $t=\frac{T}{6}$. Assume that the floor and ceiling of the box always remain horizontal.

• A. $\frac{Mg}{k}[1+\frac{{\pi }^2}{3}]$
• B. $\frac{Mg}{k}[1+\frac{{\pi }^2}{9}]$
• C. $\frac{Mg}{2k}[1+\frac{{\pi }^2}{3}]$
• D. $\frac{Mg}{2k}[1+\frac{{\pi }^2}{9}]$

Answer 1

The correct option is D $\frac{Mg}{2k}[1+\frac{{\pi }^2}{9}]$

Originally, the system keeps the spring stretched by a length $x_0$, where, $k{x}_0=2Mg$ .............(1)
As soon as the particle gets detached, it begins to fall with acceleration $g$ and the box begins to perform SHM with amplitude,
$A=\frac{2Mg}{k}-\frac{Mg}{k}=\frac{Mg}{k}$
The distance travelled (downward) by the particle in time $t$ will be, $s_1=\frac{1}{2}g{t}^2$
The distance travelled (upward) by the floor in time $t$ will be
$s_2=A-A\cos \left(\omega t\right)$, where, $\omega =\sqrt{\frac{k}{M}},t=\frac{T}{6}$ and $T=\frac{2\pi }{\omega }$
$\Rightarrow s_2=\frac{Mg}{k}-\frac{Mg}{k}\cos \left(\omega t\right)$
$\therefore H=s_1+s_2$
$=\frac{1}{2}g{t}^2+\frac{Mg}{k}-\frac{Mg}{k}\cos \left(\omega t\right)$
$=\frac{1}{2}g\frac{{\pi }^2}{9}\frac{M}{k}+\frac{Mg}{k}-\frac{Mg}{k}\cos \left(\frac{\pi }{3}\right)=\frac{Mg}{2k}[1+\frac{{\pi }^2}{9}]$