Question

The bob of a simple pendulum executes a simple harmonic motion in water with a period $\textit{t}$ , 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 $\textit{t}$ and $t_{0}$ is

• A. $t = t_{0}$
• B. $t = t_{0} / 2$
• C. $t = 2 t_{0}$
• D. $t = 4 t_{0}$

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 $=m g-(V \sigma g)=V \rho g-V \sigma g$
$V \rho \mathrm{g}^{\prime}=V \mathrm{~g}(\rho-\sigma)$, 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$.