Question

A rectangular block of mass 'm' and cross-sectional area 'A' floats on a liquid of density '\(ρ\)'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency (g=acceleration due to gravity)

• A. \(2π\sqrt{\frac{\text{m}}{\text{A}ρ\text{g}}}\)
• B. \(2π\sqrt{\frac{\text{A}ρ\text{g}}{\text{m}}}\)
• C. \(\frac{1}{2π}\sqrt{\frac{\text{A}ρ\text{g}}{\text{m}}}\)
• D. \(\frac{1}{2π}\sqrt{\frac{\text{m}}{\text{A}ρ\text{g}}}\)

Hint:

The extra buoyant force is A rho g times the displacement, like a spring of constant A rho g.

Answer 1

The Correct Option is C

Step 1: Equation of motion:
$m\ddot y = -A\rho g\,y$, so $\ddot y + \dfrac{A\rho g}{m}y = 0$.

Step 2: Compare with SHM:
This is $\ddot y + \omega^2y = 0$ with $\omega^2 = \dfrac{A\rho g}{m}$.

Step 3: Frequency:
$f = \omega/2\pi = \dfrac1{2\pi}\sqrt{\dfrac{A\rho g}{m}}$, option (C).

Final Answer:
The frequency is 1/(2 pi) times root of A rho g over m. \[ \boxed{\text{(C) }\dfrac{1}{2\pi}\sqrt{\dfrac{A\rho g}{m}}} \]