A pendulum has a period `T` for small oscillations. An obstacle is placed directly beneath the pivot, so that only the lowest one - quarter of the string can follow the pendulum bob when it swings to the left of its resting position. The pendulum is released from rest at a certain point. How long will it take to return to that point again ? In answering this question, you may assume that the angle between the moving string and the vertical stays small throughout the motion.

Options

A. `T`
B. `T//2`
C. `3T//4`
D. `T//4`

Answer

C

Explanation

Time taken the pendulum to move from `A` to `O` and from `O` to `A = (T)/(2)`
Time period of oscillation `propm sqrt(L)`
`(T_(1))/(T_(2)) = sqrt((L//4)/(L)) = (1)/(2)` or `T_(1) = (T)/(2)`
Time taken complete half the oscillation `= (T_(1))/(4)`
Total time period of oscillation` = (T)/(2) + (T)/(4) = (3T)/(4)`