Question

A uniform cylinder of length $L$ and mass $M$ having cross sectional area $A$ is suspended, with its length vertical, form a fixed point by a massless spring, such that it is half submerged in a liquid of density $\sigma$ at equilibrium position. When the cylinder is given a downward push and released, it starts oscillating vertically with a small amplitude. The time period of the oscillations of the cylinder will be:

• A. Smaller than $2\pi {\left[\frac{M}{(k+A\rho g)}\right]}^{1/2}$
• B. $2\pi \sqrt{\frac{M}{k}}$
• C. arger than $2\pi {\left[\frac{M}{(k+A\rho g)}\right]}^{1/2}$
• D. $2\pi {\left[\frac{M}{(k+A\rho g)}\right]}^{1/2}$

Answer 1

The correct option is A Smaller than $2\pi {\left[\frac{M}{(k+A\rho g)}\right]}^{1/2}$

$\text{Solution}:$

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 $F=-(kx+Ax\rho g)$

or $a=-(\frac{k+\rho Ag}{M})x$

Now, $f=\frac{1}{2\pi }\sqrt{|\frac{a}{x}}|$

$=\frac{1}{2\pi }\sqrt{\frac{k+\rho Ag}{M}}$

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

$\text{Hence A is the correct option}$