Question

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 $\rho$ at equilibrium position. When the cylinder is given a small downward push and 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.
• B. $\frac{1}{2\pi }(\frac{k+A\rho g}{M}{)}^{\frac{1}{2}}$
• C. $\frac{1}{2\pi }(\frac{k+\rho g{L}^2}{M}{)}^{\frac{1}{2}}$
• D. $\frac{1}{2\pi }(\frac{k+A\rho g}{Apg}{)}^{\frac{1}{2}}$

Answer 1

The correct option is B

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

When cylinder is displaced by an amount x from its mean position, spring force and upthrust both will increase. Hence,

Net restoring force = extra spring force + extra upthrust

Or $F=-(kx+Ax\rho g)ora=-(\frac{k+\rho Ag}{M})x$

Now, $f=\frac{1}{2\pi }\sqrt{|\frac{a}{x}}|=\frac{1}{2\pi }\sqrt{\frac{k+\rho Ag}{M}}$