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The bob of a simple pendulum executes simple harmonic motion in water with a period $t$, while the period of oscillation of the bob is ${t_0}$ in air. Neglecting frictional force of water and given that the density of the bob is $(4/3) ×1000 kg/m^3$. What relationship between $t$ and ${t_0}$ is true
$t = {t_0}$
$t = {t_0}/2$
$t = 2{t_0}$
$t = 4{t_0}$
Solution
(c) $\because \,\,{t_o} = 2\,\pi \sqrt {\frac{l}{g}} $
Effective weight of bob inside water,
$W' = mg – {\rm{thrust}} = V\rho g – V\rho 'g$
$ \Rightarrow V\,\,\rho {g_{eff}} = V(\rho – \rho ')g,$ where, $\rho $ = Density of bob
$ \Rightarrow {g_{eff}} = \left( {1 – \frac{{\rho '}}{\rho }} \right)\,g$ and $\rho '$ = Density of water
$\therefore t = 2\,\pi \sqrt {\frac{l}{{{g_{eff}}}}} = 2\,\pi \sqrt {\frac{l}{{(1 – \rho '/\rho )g}}} $ $ (\because \rho ' = {10^3}kg/{m^3} \,\,\rho = \frac{4}{3} \times {10^3}kg/{m^3}) $
$\therefore \frac{t}{{{t_0}}} = \sqrt {\frac{1}{{1 – \rho '/\rho }}} = \sqrt {\frac{1}{{1 – \frac{3}{4}}}} $
$ \Rightarrow t = 2\,{t_0}$.
