Question

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 its density of the bob is $\left(\frac{4}{3}\right)\times 1000kg/m^3$. What relationship between $t$ and $t_0$ is true?

• A. $t=t_0$
• B. $t=t_0/2$
• C. $t=2t_0$
• D. $t=4t_0$

Answer 1

Answer: (C)

$t=2t_0$

$\because t_o=2\pi \sqrt{\frac{l}{g}}$
Effective weight of bob inside water,
$W^{{}^{\prime }}=mg-thrust=V\rho g-V{\rho }^{{}^{\prime }}g$
$\Rightarrow V\rho geff=V(\rho -{\rho }^{{}^{\prime }})g$, where $\rho =$ Density of bob
$\Rightarrow geff=(1-\frac{{\rho }^{{}^{\prime }}}{\rho })g$ and ${\rho }^{{}^{\prime }}=$ Density of water

$\therefore t=2\pi \sqrt{\frac{l}{geff}}=2\pi \sqrt{\frac{l}{(1-{\rho }^{{}^{\prime }}/\rho )g}}$

$\therefore \frac{t}{t_0}=\sqrt{\frac{1}{(1-{\rho }^{{}^{\prime }}/\rho )}}=\sqrt{\frac{1}{1-\frac{3}{4}}}$ ($\therefore {\rho }^{{}^{\prime }}={10}^{3}kg/m^3\rho =\frac{4}{3}\times {10}^{3}kg/m^3$)
$\Rightarrow t=2t_0$.