A wooden cylinder of mass 20 g and area of cross-section 1 cm 2 , having a piece of lead of mass 60 g attached to its bottom, floats in water. The cylinder is depressed and then released. The frequenc

Question

A wooden cylinder of mass 20 g and area of cross-section 1 cm 2 , having a piece of lead of mass 60 g attached to its bottom, floats in water. The cylinder is depressed and then released. The frequency of oscillations is N π s − 1 . Find the value of N . [Neglect the volume of water displaced by the lead piece, take g = 9.8 m/s 2 , density of water ρ w = 1 g cm - 3 ]

TestHub · Accepted Answer

Suppose that the loaded wooden block sinks upto a height h. Then,

weight of water displaced by the block

= weight of the block with lead

If a is area of cross-section of the block and σ, the density of water, then

$a h σ g = (M + m) g$

$or h = \frac{M + m}{a σ}$

$= \frac{20 + 60}{1 \times 1} = 80 cm (∵ density of water = 1 g {cm}^{- 3})$

Now for small displacement ' $x$ ' from equilibrium

$σ g a (x + h) - (M + m) g = -(M + m) A$

$σ g a x = -(M + m) A$

$A=- \frac{σ g a}{M + m} x$

$f = \frac{1}{2 π} \sqrt{\frac{g}{h}}$

$= \frac{1}{2 π} \sqrt{\frac{980}{80}}$ $= \frac{1.75}{π} s^{- 1}$

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