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A pendulum has period T for small oscillations. An obstacle is placed directly beneath the pivot, son that only the lowest one quarter of the string can follow the pendulum ob when it swings in the left of its resting position as shown in the figure. The pendulum is released from rest at a certain point A. The time taken by its to return to that point is:

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$\frac{T}{2}$

$\frac{3 T}{4}$

$\frac{T}{4}$

answer is C.

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Detailed Solution

Time taken the pendulum to move from A to O and from O to A = $\frac{T}{2}$
Time period of oscillation $\propto \sqrt{L}$
$\frac{T_{1}}{T} = \sqrt{\frac{\frac{L}{4}}{L}} = \frac{1}{2} or T_{1} = \frac{T}{2}$

Time taken to complete half the oscillation = $\frac{T_{1}}{4}$
Total time period of oscillation = $\frac{T}{2} + \frac{T}{4} = \frac{3 T}{4}$

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