A cylindrical container having area of cross-section equal to 4A contains a non- viscous liquid of density $2 ρ$ .A cylindrical block of wood having cross sectional area A and length L has density $ρ$ , held vertically with its lower surface touching the liquid. Now it is released from this position. Assume that the depth of the container is sufficient and the cylinder does not touch the bottom.
The amplitude of oscillation of the wooden cylinder is $\frac{3 L}{n}$ .Then $n$ is____

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answer is 8.

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Detailed Solution

When the cylinder moves down by x, the liquid level rises by y.
$3 A y = A x \Rightarrow y = \frac{x}{3}$
Since volume of liquid is constant
Buoyancy force on cylinder
$= (\frac{4 x}{3} A) (2 ρ) g$

In equilibrium $= \frac{8 x_{0}}{3} A ρ g = A L ρ g \Rightarrow x_{0} = \frac{3 L}{8}$
Since, equilibrium position, is $x_{0}$ below the original position (where, cylinder was at rest),Hence amplitude is $A = x_{0} = \frac{3 L}{8}$