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Q.

A small ball is suspended by a thread of length l = 1 m at the point O on the wall, forming a small angle α=2∘ with the vertical (as shown in figure). Then the thread with ball was deviated through a small angle β=4∘ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, the oscillation period of such a pendulum (in seconds) is ___________ (Take g=π2).

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answer is 1.33.

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Detailed Solution

The time period of free, oscillation of pendulum T=2πlg=2sTime taken by bob to go from extreme position A to mean position B is =T4Time taken by bob to move from mean position B to position C (where its angular displacement α is half the angular amplitude β) is found from equation α=βsin⁡2πTt.Solving we get t=T12⇒ Total time period of oscillation=2T4+T12=23T=43s
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