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 $\left(4/3\right)\times 1000kg/m^3$. 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

The correct option is C $t=2t_0$

Let $V$ be the volume of the bob.

Thus mass of the bob $M={\rho }_{b}V=\frac{4}{3}\times 1000Vkg$

Acceleration due to gravity in air is $g.$

Effective weight of the bob in water $M{g}^{\prime }=Mg-B$ where buoyant force $B=\left(V{\rho }_{water}\right)g$

$\frac{4}{3}\times 1000V{g}^{\prime }=\frac{4}{3}\times 1000Vg-V\times 1000g$ (as ${\rho }_{water}=1000kg/m^3$)

$\frac{4}{3}\times 1000V{g}^{\prime }=\frac{1}{3}\times 1000Vg$

$⟹g^{\prime }=\frac{g}{4}$

Now time period of the bob in air $t_o=2\pi \sqrt{\frac{l}{g}}$

Time period of the bob in water $t=2\pi \sqrt{\frac{l}{g^{\prime }}}$ $=2\pi \sqrt{\frac{4l}{g}}$ $=2\times 2\pi \sqrt{\frac{l}{g}}$

$⟹t=2t_o$