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. 2 m T A g π ρ = where m is mass of the body and ρ is density of the liquid.

This is a long answer type question as classified in NCERT Exemplar

Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position

So mg= buoyant forces = $\rho A{x}_{o}g$

When it is displaced by further displacement x, the buoyant force is A (x_o+x) $\rho g$

Net restoring force = buoyant forces -weight

=A (x_o+x) $\rho g$ -mg

=A $\rho gx$

As displacement x is downward and restoring force is upward

F_{restoring =}-A $\rho gx$ =-kx

So motion is SHM

Acceleration a=F_restoring/m=-kx/m

a=-w²x

w²=k/m

w= $\sqrt[]{\frac{k}{m}}$

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

<div><div><p><span data-teams="true">This is a long answer type question as classified in NCERT Exemplar</span></p><p>Let the log be passed and the vertical displacement at the vertical displacement at the equilibrium position</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745560231php9acUVK_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745560231php9acUVK.jpeg" alt="" width="243" height="133"></picture></div><div><p>So mg= buoyant forces =<span contenteditable="false"> <math> <mi>ρ</mi> <mi>A</mi> <msub> <mrow> <mrow> <mi>x</mi> </mrow> </mrow> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> </msub> <mi>g</mi> </math> </span></p><p>When it is displaced by further displacement x, the buoyant force is A (x_o+x)<span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> </math> </span></p><p>Net restoring force = buoyant forces -weight</p><p>=A (x_o+x)<span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> </math> </span> -mg</p><p>=A <span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <mi>x</mi> </math> </span></p><p>As displacement x is downward and restoring force is upward</p><p>F_restoring₌-A<span contenteditable="false"> <math> <mi>ρ</mi> <mi>g</mi> <mi>x</mi> </math> </span> =-kx</p><p>So motion is SHM</p><p>Acceleration a=F_restoring/m=-kx/m</p><p>a=-w²x</p><p>w²=k/m</p><p>w=<span contenteditable="false"> <math> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p><p>T= 2<span contenteditable="false"> <math> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> <mo>=</mo> <mn>2</mn> <mi>π</mi> <mroot> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> <mi>ρ</mi> <mi>g</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> <mrow></mrow> </mroot> </math> </span></p></div></div></div></div>

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