# A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical…

Q: A uniform cylinder of length L and mass M having cross-sectional area A is suspended, with its length vertical, from a fixed point by a massless spring, such that it is half-submerged in a liquid of density ρ at equilibrium position. When the cylinder is given a small downward push and released to 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) $\large \frac{1}{2\pi} (\frac{k – A \rho g}{M})^{1/2}$

(b) $\large \frac{1}{2\pi} (\frac{k + A \rho g}{M})^{1/2}$

(c) $\large \frac{1}{2\pi} (\frac{k + \rho g L^2}{M})^{1/2}$

(d) $\large \frac{1}{2\pi} (\frac{k + A \rho g}{A \rho g})^{1/2}$

Ans: (b)

Sol: Let cylinder is displaced by an amount x from its mean position .

The net restoring force , $\large F = -(k x + A x \rho g)$

$\large M a = -(k x + A x \rho g)$

$\large a = -\frac{(k + A \rho g)}{M} x$

In S.H.M , a = -ω2 x

Hence , $\large \omega^2 = \frac{(k + A \rho g)}{M}$

$\large \omega = \sqrt{\frac{(k + A \rho g)}{M}}$

Frequency of oscillation $\large f = \frac{\omega}{2\pi}$

$\large f = \frac{1}{2\pi}\sqrt{\frac{(k + A \rho g)}{M}}$