KVPY SX SB Exam 2012 Question Paper

Question : 100 of 120

Marks: +1, -0

A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle θ with the vertical, the angle ϕ which the acceleration vector of the bob makes with the string is given by

$ϕ=0$
$ϕ=\tan^{-1}({\tanθ}/2)$
$ϕ=\tan^{-1}({2\tanθ} )$
$ϕ=π/2$

Validate

Solution:

By energy conservation

\tan ϕ = a_t/a_c

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