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+1 vote
5.0k views
in Physics by (75.1k points)

The bob of a simple pendulum having length 'ℓ' is displaced from its equilibrium position by an angle of θ and released. If the velocity of the bob, while passing through its equilibrium position is v, then v = …….. 

(A) √{2gℓ(1 – cos θ)} 

(B) √{2gℓ(1 + sin θ)} 

(C) √{2gℓ(1 – sin θ)} 

(D) √{2gℓ(1 + cos θ)}

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1 Answer

+1 vote
by (70.5k points)

The correct option (A) √{2gℓ(1 – cos θ)}

Explanation:

cos θ = (h/ℓ) i.e. h = ℓ cos θ 

∴ ℓ – h = ℓ – ℓ cos θ = ℓ (1 – cos θ) (1) 

KE at 0 = PE at A 

(1/2) mV2 = mgh' 

V2 = 2gh' 

here h' = ℓ (1 – cos θ) ----- from (1) 

∴ V2 = 2gℓ (1 – cos θ) V = √{2gℓ(1 – cos θ)}

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