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

Question

The bob of simple pendulum having length l, is displaced from mean position to an angular position q with respect to vertical. If it is released, then velocity of bob at equilibrium position (A)  √ 2 gl ( 1 - cos θ)  (B)  √ 2 gl ( 1 + cos θ)  (C)  √ 2 gl cos θ  (D)  √ 2gl

Tardigrade Admin · Accepted Answer

In triangle O A C, cos θ=A / l or, O A=l cos θ ∴ A B=l(1- cos θ)=h At point, C the velocity of bob =0. The vertical acceleration =g ∴ v2=2 g h or, v=√2 g l(1- cos θ)

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

$2 g l (1 - cos θ)$

$2 g l (1 + cos θ)$

$2 g l cos θ$

$2 g l$

In $△ O A C, cos θ = A / l$
or, $O A = l cos θ$
$∴ A B = l (1 - cos θ) = h$
At point, $C$ the velocity of bob $= 0$ .
The vertical acceleration $= g$
$∴ v^{2} = 2 g h$
or, $v = 2 g l (1 - cos θ)$

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

2gl(1−cosθ)​

2gl(1+cosθ)​

2glcosθ​

2gl​