Answer:
Let
P represent the equilibrium position of a cylinder when floating in a liquid as
shown in.
Let
\[m=\]mass of cross-section of the cylinder,
\[\rho
=\]density of the liquid
\[l=\]length
of the cylinder dipping in the liquid when in equilibrium position.
If \[{{F}_{1}}\] is the upward thrust, then according to
Archimedes principle,
\[{{F}_{1}}=\]weight
of the liquid displaced by the length \[l\] of
the cylinder
\[=(Al)\rho
g=Al\rho g\]
As the cylinder is
in equilibrium,
\[{{F}_{1}}=mg\]
or \[mg=Al\rho
g\]
or \[m=Al\rho
\] ?(i)
let
the cylinder be pushed down through a certain distance \[y\] as shown by its position \[Q\]. If \[{{F}_{2}}\] is the upward thrust,
\[{{F}_{2}}=\]weight
of the liquid displaced by the length \[(l+y)\] of
the cylinder,
\[=A(l+y)\rho
g\]
Restoring force
acting on the cylinder, i.e.,
\[F={{F}_{2}}-mg={{F}_{2}}-{{F}_{1}}=A(l+y)\rho
g-Al\rho g=Ay\rho g\] or \[F=-A\rho gy\] ?(ii)
(negative
sign indicates that F is opposite to y)
If \[{{d}^{2}}y/d{{t}^{2}}\] is the acceleration produced in the
cylinder,
\[F=m({{d}^{2}}y/d{{t}^{2}})\] ?(iii)
From
eqns. (i), (ii) and (iii),
\[(Al\rho
){{d}^{2}}y/d{{t}^{2}}=-A\rho gy\]
or \[\frac{{{d}^{2}}y}{d{{t}^{2}}}=-\frac{g}{l}y=-{{\omega
}^{2}}y\]
or \[\frac{{{d}^{2}}y}{d{{t}^{2}}}+{{\omega
}^{2}}y=0\] ?(iv)
where,
\[{{\omega }^{2}}=g/l\]
or \[\omega
=\sqrt{g/l}\]
Eqn. (iv) represents
as SHM. Thus, the cylinder executes SHM and its time period is given by
\[T=\frac{2\pi
}{\omega }=2\pi \sqrt{\frac{l}{g}}\] ?(v)
From eqns. (i) and
(v)
\[T=2\pi
\sqrt{\frac{m/A\rho }{g}}=2\pi \sqrt{\frac{m}{A\rho g}}\]
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