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 $ (4/3)\times 1000kg/{{m}^{3}} $ . What relationship between $ t $ and $ {{t}_{0}} $ is true?

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

Answer 1

Answer: (C)

The time period of simple pendulum in air $ T={{t}_{.0}}=2\pi \sqrt{\left( \frac{l}{g} \right)} $ ?. (i) $ l $ ,being the length of simple pendulum. In water, effective weight of bob $ w= $ weight of bob in air - upthrust $ \Rightarrow $ $ \rho V{{g}_{eff}}=mg-mg $ $ =\rho Vg-\rho Vg=(\rho -\rho )Vg $ where $ \rho = $ density of bob, $ \rho = $ density of water $ \therefore $ $ {{g}_{eff}}=\left( \frac{\rho -\rho }{\rho } \right)g=\left( 1-\frac{\rho }{\rho } \right)g $ $ \therefore $ $ t=2\pi \sqrt{\left[ \frac{l}{\left( 1-\frac{\rho }{\rho } \right)g} \right]} $ ? (ii) Thus, $ \frac{t}{{{t}_{0}}}=\sqrt{\left( \frac{1}{\left( 1-\frac{\rho }{\rho } \right)} \right)} $ $ =\sqrt{\left( \frac{1}{1-\frac{1000}{(4/3)\times 1000}} \right)}=\sqrt{\left( \frac{4}{4-3} \right)} $ $ =2\Rightarrow t=2{{t}_{0}} $