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A cylindrical log of wood of height h and area of cross-section A floats in water. It is pressed and then released. Show that the log would execute S.H.M. with a time period.
$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$

where m is the mass of the body and ρ is the density of the liquid.

Answers (1)

If we press a log downward into the liquid, a buoyant force acts on it, and due to inertia it moves upwards from its mean position & comes down again due to gravity.

Thus, the restoring force on the block = Buoyant force (B.F.) – mg

The volume of liquid displaces by block = V

When it floats,

mg = BF

Or, $mg = V\rho g$

$mg = Ax_{0} \rho g$ …….. (i)

Area of crosssection = A

Height of liquid block = $x_{0}$

When pressed in water, the total height of the block in water = $(x + x_{0})$

Thus, net restoring force = $[A(x + x_{0})]\rho g -mg$

Frest $= -Ax\rho g$ …….. ( since BF is upward & x is downward)

Frest proportional to x

Hence, the motion here is SHM.

Now, $kA\rho g$

$a = -\omega ^{2}x\omega$

$= \frac{k}{m}$

Thus, $T= 2 \pi \sqrt{\frac{m}{k}}$

Frest $= -A\rho gx$

$ma= -A\rho gx$

$a=\frac{ -A\rho gx}{m}$

$-\omega ^{2}x=\frac{ -A\rho gx}{m}$

$K = A\rho g$

$\left (\frac{2\pi 2}{T} \right )= \frac{A\rho g}{m}$

Thus,

$\frac{T}{2\pi} = \frac{m}{A\rho g}$

Thus,

$T = 2 \pi \sqrt{\frac{m}{A \rho g}}$

Posted by

infoexpert24

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