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Question

# A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle θ with the vertical, then the angle α, which the acceleration vector of the bob makes with the string is equal to

A
0
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B
tan1(2tanθ)
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C
tan1(tanθ2)
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D
π/2
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Solution

## The correct option is B tan−1(tanθ2)On applying energy conservation taking the bottom line as reference, Potential Energy at top = (Potential Energy + Kinetic Energy) at given instant mgl=mgl(1−cosθ)+12mv2mglcosθ=mv22So v=√2glcosθtangential acceleration of bob is aT=gsinθ Radial acceleration aR=v2l=2gcosθacceleration vector angle with string is tanα=aTaR=gsinθ2gcosθα=tan−1(tanθ2)

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