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 oscillations of the cylinder is?

• A. $\frac{1}{2\pi }{\left(\frac{k-A\rho g}{M}\right)}^{1/2}$
• B. $\frac{1}{2\pi }$ ${\left(\frac{k+A\rho g}{M}\right)}^{1/2}$
• C. $\frac{1}{2\pi }{\left(\frac{k+\rho gL}{M}\right)}^{1/2}$
• D. $\frac{1}{2\pi }{\left(\frac{k+A\rho g}{A\rho g}\right)}^{1/2}$

Answer 1

Answer: (B)

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

When the cylinder is pushed and released then, the force applied and the up thrust force will be equal and it can be written as,

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

The acceleration is given as,

$a=-\frac{\left(kx+Ax\rho g\right)}{M}$

The frequency of the oscillation is given as,

$f=\frac{1}{2\pi }\sqrt{\left|\frac{a}{x}\right|}$

$=\frac{1}{2\pi }\sqrt{\frac{\left(k+A\rho g\right)}{M}}Hz$

Thus, the frequency of the oscillation of the cylinder is $\frac{1}{2\pi }\sqrt{\frac{\left(k+A\rho g\right)}{M}}Hz$.