question_answer 62)

                  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 mass of the body and p is density of the liquid.

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)                 \[T=2\pi \sqrt{\frac{m/A\rho }{g}}=2\pi \sqrt{\frac{m}{A\rho g}}\]