Q. 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 ($\frac{4}{3} \times 1000\ kg - m^{3}$). What relationship between $t\ and\ t_{0}$ is true?
Explanation
(c) The time period of simple pendulum in air
$T = t_{0} = 2\pi\sqrt{\left( \frac{l}{g} \right)}\ $ …(i)
$l$, being the length of simple pendulum.
In water, effective weight of bob
$w^{'} =$ weight of bob in air$\ –\ $upthrust
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \rho Vg_{eff} = mg - m'g$
$= \rho Vg - \rho^{'}Vg = \left( \rho - \rho^{'} \right)Vg$
where $\rho^{'} =$ density of bob,
$\rho = \ $ density of water
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ g_{eff} = \left( \frac{\rho - \rho'}{\rho} \right)g = \left( 1 - \frac{\rho'}{\rho} \right)g$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 2\pi\sqrt{\left\lbrack \frac{l}{\left( 1 - \frac{\rho'}{\rho} \right)g} \right\rbrack}$ …(ii)
Thus, $\frac{t}{t_{0}} = \sqrt{\left\lbrack \frac{1}{\left( 1 - \frac{\rho'}{\rho} \right)} \right\rbrack}$
$= \sqrt{\left( \frac{1}{1 - \frac{1000}{(\frac{4}{3} \times 1000)}} \right)} = \sqrt{\left( \frac{4}{4 - 3} \right)} = 2$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ t = 2t_{0}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.