Q. A uniform cylinder of length L and mass M having cross sectional area A is suspended with its vertical length, from a fixed point by a massless spring, such that it is half submerged in a liquid of density d at equilibrium position. When released, it starts oscillating vertically with a small amplitude. If the force constant of the spring is k, the frequency of oscillation of the cylinder is
A) $\frac{1}{2\pi}\left( \frac{k - A\ d\ g}{M} \right)^{1/2}$
B) $\frac{1}{2\pi}\left( \frac{k + A\ d\ g}{M} \right)^{1/2}$
C) $\frac{1}{2\pi}\left( \frac{k - d\ g\ L}{M} \right)^{1/2}$
D) $\frac{1}{2\pi}\left( \frac{k + A\ g\ L}{A\ d\ g} \right)^{1/2}$
Explanation
(b) When the cylinder is given a small downward displacement, say $y$, the additional restoring force is due to (i) additional extension $y$, which is
$F_{1} = ky$ (ii) Additional buoyancy, which is $F_{2} = A\ Y\ d\ g$
Total restoring force, $- F = F_{1} + F_{2} = (k + Adg)_{y}$
=new force constant
$\therefore\ n - \frac{1}{2\pi}\sqrt{\frac{k^{'}}{k}} = \frac{1}{2\pi}\sqrt{\frac{k + Adg}{M}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.