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The bob of a simple pendulum having length l is displaced from the mean position to an angular position θ with respect to the vertical. If it is released, then the velocity of the bob at the equilibrium position is

Asked in CBSE AIPMT 2000 · Kinetic, potential and total energy

• (1) √2gl(1-cos θ)
• (2) √2gl(1+cos θ)
• (3) √2gl cos θ
• (4) √2gl

Step-by-step solution

Why the other options are wrong

• (2) comes from writes the drop as l(1+cos θ); that would be the fall from θ down past the lowest point to the same height on the far side of the pivot, which the bob never reaches
• (3) comes from takes the drop as l cos θ, which is the bob's height *below the pivot* at the release point, not the height it falls
• (4) comes from takes the drop as the full length l, which would need a release from the horizontal, θ=90°

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