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 p 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

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 p 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) (1/2π)((k-Aρ g/M)) (B) (1/2π)((k+Aρ g/M))1/2 (C) (1/2π)((k+ρ gL2/M))1/2 (D) (1/2π)((k+Aρ g/Aρ g))1/2

Tardigrade Admin · Accepted Answer

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 = - (k x + A x ρ g)

or

a = - (\frac{k + ρ A g}{M}) x

Now,

f = \frac{1}{2 π} ∣ ∣ \frac{a}{x} ∣ ∣

= \frac{1}{2 π} \frac{k + ρ A g}{M}

∴

Correct optionis (b)

$\frac{1}{2 π} (\frac{k - A ρ g}{M})$

$\frac{1}{2 π} (\frac{k + A ρ g}{M})^{1/2}$

$\frac{1}{2 π} (\frac{k + ρ g L ^{2}}{M})^{1/2}$

$\frac{1}{2 π} (\frac{k + A ρ g}{A ρ g})^{1/2}$

2π1​(Mk−Aρg​)

2π1​(Mk+Aρg​)1/2

2π1​(Mk+ρgL2​)1/2

2π1​(Aρgk+Aρg​)1/2

$\frac{1}{2 π} (\frac{k - A ρ g}{M})$

$\frac{1}{2 π} (\frac{k + A ρ g}{M})^{1/2}$

$\frac{1}{2 π} (\frac{k + ρ g L ^{2}}{M})^{1/2}$

$\frac{1}{2 π} (\frac{k + A ρ g}{A ρ g})^{1/2}$