question_answer

Determine the period of small oscillations of a mathematical pendulum, that is a ball suspended by a thread l = 20 cm in length, if it is located in a liquid whose density is three times less than that of the ball. The resistance of the liquid is to be neglected.

A)  2.2s                

B) 1.1s

C) 0.5s                 

D) 3.1s

Correct Answer: B

Solution :

[b]: Let us depict the forces acting on the oscillating ball at an arbitrary angular position \[\theta \], relative to equilibrium position where \[{{F}_{B}}\]is the force of buoyancy.        The equation of motion for ball \[-mgl\sin \theta +{{F}_{B}}l\sin \theta =m{{l}^{2}}\ddot{\theta }\]                   ...(i) Using \[m=\frac{4}{3}\pi {{r}^{3}}\sigma ,{{F}_{B}}=\frac{4}{3}\pi {{r}^{3}}\rho g\]and \[\theta \simeq \theta \]for small \[\theta \], in equation (i), we get\[\ddot{\theta }=-\frac{g}{l}\left( 1-\frac{\rho }{\sigma } \right)\theta \] Thus the time period of the ball \[T=2\pi \frac{1}{\sqrt{\frac{g}{l}\left( 1-\frac{\rho }{\sigma } \right)}}=2\pi \sqrt{\frac{l/g}{1-\frac{1}{3}}}\] \[=6.28\sqrt{\frac{0.2/9.8}{2/3}}=1.1s\]