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 \[\left( 4 / 3 \right) \times  1000 kg/{{m}^{3}}\]. What relationship between t and \[{{t}_{0}}\] is true

A) \[\operatorname{t}=2t\Omega \]

B)        \[\operatorname{t}={{t}_{0}}/2\]

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

D)        \[t=4{{t}_{0}}\]

Correct Answer: A

Solution :

\[\operatorname{t}=2\pi \sqrt{\frac{\ell }{{{g}_{eff}}}};\,\,{{t}_{0}}=2\pi \sqrt{\frac{\ell }{g}}\] \[\operatorname{Net}\,\,force=\,\,\left( \frac{4}{3}-1 \right)\times 1000\,Vg\,=\frac{1000}{3}Vg\] \[{{g}_{eff}}\,\,=\,\,\frac{1000\,Vg}{3\times \frac{4}{3}\times 1000\,V}\,\,=\,\,\frac{g}{4}\] \[\therefore \,\,t=2\pi \sqrt{\frac{\ell }{g/4}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t=2{{t}_{0}}\]