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The bob of simple pendulum having length 1, is displaced from mean position to an angular position o with respect to vertical. If it is released, then velocity of bob at equilibrium position is |
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Answer» `sqrt(2g(1- cos theta))` `therefore AB = 1(1- cos theta) =h` At point C, the velocity of bob =0. The vertical ACCELERATION =G `therefore v^(2) = 2gh` or, `v=sqrt(2gl(1- cos theta))` 
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