Question

A cylindrical plastic bottle of negligible mass is filled with $310ml$ of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $\omega$. If the radius of the bottle is $2.5cm$ then $\omega$ is close to:
(density of water $={10}^{3}kg/m^3$)

• A. $3.75rad$ $s^{-1}$
• B. $1.25rad$ $s^{-1}$
• C. $2.50rad$ $s^{-1}$
• D. $2.5\sqrt{10}rad{s}^{-1}$

Answer 1

The correct option is D $2.5\sqrt{10}rad{s}^{-1}$

$\begin{array}{c}EquationofS.H.M\\ a=-\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x\\ \omega =\sqrt{\frac{k}{m}}\\ wherex=A\sin \left(\omega t+\delta \right)\\ A\times \rho g=F_{rest}\Rightarrow \left(\pi r^2\rho g\right)x=F_{rest}\\ {\omega }^2=\frac{\pi r^2\rho g}{m}\Rightarrow \omega =\sqrt{\frac{\pi g}{v}}\\ \sin ce,m=\rho V\\ \omega =2.5\times {10}^{-2}\sqrt{\frac{3.14\times 10}{310\times {10}^{-6}}}=2.5\sqrt{10}rad{s}^{-1}\\ Hence,theoptionDisthecorrectanswer.\end{array}$