Question

The bob of a simple pendulum executes a simple harmonic motion in water with a period $\text{t}$ , 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 $(4/3)\times 1000\mathrm{kg}{\mathrm{m}}^{-3}$. The correct relationship between $\text{t}$ and $t_0$ is

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

Answer 1

Answer: (C)

$t_0=2\pi \sqrt{l/g}$
Due to upthrust of water on the top, its apparent weight decreases upthrust $=$ weight of liquid displaced
$\therefore$ Effective weight $=mg-(V\sigma g)=V\rho g-V\sigma g$
$V\rho {\mathrm{g}}^{\prime }=V\mathrm{g}(\rho -\sigma )$, where
$\sigma$ is density of water
or ${\mathrm{g}}^{\prime }=\mathrm{g}\left(\frac{\rho -\sigma }{\rho }\right)$
$\therefore t=2\pi \sqrt{l/{\mathrm{g}}^{\prime }}=2\pi \sqrt{\frac{l\rho }{\mathrm{g}(\rho -\sigma )}}$
$\therefore \frac{t}{t_0}=\sqrt{\frac{l\rho }{\mathrm{g}(\rho -\sigma )}\times \frac{\mathrm{g}}{l}}=\sqrt{\frac{\rho }{\rho -\sigma }}$
$=\sqrt{\frac{4\times 1000/3}{\left(\frac{4000}{3}-1000\right)}}=2$ $or$ t=t_0 \times 2=2 t_0$.