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 the 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_o
• B. t=t_o/2
• C. t=2t_o
• D. t=4t_o

Answer 1

The correct option is C t=2t_o

$\therefore t_0=2\pi \sqrt{\frac{1}{g}}$
Effective weight of bob inside water,
$W^{\prime }=mg-thrust=V\rho g-V{\rho }^{\prime }g$
$\Rightarrow V\rho g_{eff}=V(\rho -{\rho }^{\prime })g,$ where, $\rho$=Density of bob
$\Rightarrow g_{eff}=(1-\frac{{\rho }^{\prime }}{\rho })g$ and ${\rho }^{\prime }$=Density of water
$\therefore t=2\pi \sqrt{\frac{1}{g_{eff}}}=2\pi \sqrt{\frac{1}{(1-{\rho }^{\prime }/\rho )g}}\therefore \frac{t}{t_0}=\sqrt{\frac{1}{1-{\rho }^{\prime }/\rho }}=\sqrt{\frac{1}{1-\frac{3}{4}}}$
$(\because {\rho }^{\prime }={10}^{3}kg/m^3$
$\rho =\frac{4}{3}\times {10}^{3}kg/m^3$
$\Rightarrow t=2t_0.$