Question

A plank of mass $m=9\text{kg}$ and area of cross-section $A=3{\text{cm}}^2$ connected to a spring of spring constant $k=6\text{N/m}$ is floating in a liquid of density $\rho =1000{\text{kg/m}}^3$ as shown in figure. When slightly displaced from the mean position, the plank starts making simple harmonic oscillations. Considering $g=10{\text{m/s}}^2$, the time period of these oscillations will be

• A. $2\pi \text{sec}$
• B. $\pi \text{sec}$
• C. $4\pi \text{sec}$
• D. $3\pi \text{sec}$

Answer 1

The correct option is A $2\pi \text{sec}$

Let the mass be displaced by $x$ in downward direction.


From the FBD,
Equation of motion can be written as, $m\ddot{x}=-kx-\rho Axg$
$\Rightarrow \ddot{x}=-\frac{(k+\rho Ag)}{m}x$
Comparing the above equation with $a=-{\omega }^{2}x$, we get
$\omega =\sqrt{\frac{k+\rho Ag}{m}}$
From this,
Time period of oscillation $\left(T\right)=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k+\rho Ag}}$
From the data given in the question, substituting the values of $m,k,\rho ,A$ and $g$, we have
$T=2\pi \sqrt{\frac{9}{6+1000\times 3\times {10}^{-4}\times 10}}$
$\Rightarrow T=2\pi \text{sec}$
Thus, option (a) is the correct answer.