First slide

Vertical circular motion

Question

A pendulum is oscillating in a vertical plane. The acceleration at the extreme position and lowest position are equal. Then the angular amplitude of the pendulum is

Moderate

A

$\tan^{- 1} (2)$
B

$\tan^{- 1} (\sqrt{2})$
C

$\tan^{- 1} (1 / 2)$
D

$2 \tan^{- 1} (\frac{1}{2})$

Solution

acceleration at the extreme position gsin $θ$ and lowest position $\frac{v^{2}}{l}$ .
$gsinθ = \frac{v^{2}}{l}$
From conservation of energy $\frac{1}{2} {mv}^{2} = mgl (1 - cosθ)$
$\frac{{mv}^{2}}{l} = 2 mg (1 - cosθ)$
$gsinθ = 2 g (1 - cosθ)$
$sinθ = 4 \sin^{2} (\frac{θ}{2})$
$2 \sin \frac{θ}{2} \cos \frac{θ}{2} = 4 \sin^{2} \frac{θ}{2}$
$\tan \frac{θ}{2} = \frac{1}{2}$
$\frac{θ}{2} = \tan^{- 1} (\frac{1}{2})$
$θ = 2 \tan^{- 1} \frac{1}{2}$

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