question_answer

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) \[t=4{{t}_{0}}\]

Correct Answer: C

Solution :

[c] \[{{t}_{0}}=2\pi \sqrt{\frac{\ell }{g}}\]. The restoring force in a liquid  \[F=-(mg-V{{\rho }_{e}}g)\sin \theta \] \[=-\,\left( mg-\frac{m}{\left( \frac{4}{3}\times 1000 \right)}\times 1000g \right)\left( \frac{x}{\ell } \right)\] \[or\,\,\,a=\left( g-\frac{3g}{4} \right)\left( \frac{-x}{\ell } \right)=\frac{g}{4}\left( \frac{-x}{\ell } \right)\] \[\therefore \,\,t=2\pi \sqrt{\frac{\ell }{(g/4)}}=2{{t}_{0}}\].