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MHT CET 2024 (16 May, Shift 2) · Simple Pendulum

Simple Pendulum question from MHT CET 2024 (16 May, Shift 2)

Q37MHT CET 2024 · 16 MAY · SHIFT 2Medium

Simple Pendulum

The bob of a pendulum of length ' $l$ ' is pulled aside from its equilibrium position through an angle ' $\theta$ ' and then released. The bob will then pass through its equilibrium position with speed ' $v$ ', where ' $v$ ' equal to ( $g =$ acceleration due to gravity)

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Answer: (a)

Height fallen from the extreme to the mean position: $h = l - l\cos\theta = l(1-\cos\theta)$ .
Energy conservation: $\frac{1}{2}mv^2 = mgl(1-\cos\theta)$ .
So the answer is (a) $v = \sqrt{2gl(1-\cos\theta)}$ .

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