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Oscillations question

2000 · Q162
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Oscillations question

2000 · Q162

NEETPhysicsOscillationsMCQ+4 / −1
The bob of simple pendulum having length lll, 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
  1. A
    2gl(1−cos⁡θ)\sqrt {2gl\left( {1 - \cos \theta } \right)}2gl(1−cosθ)​
  2. B
    2gl(1+cos⁡θ)\sqrt {2gl\left( {1 + \cos \theta } \right)}2gl(1+cosθ)​
  3. C
    2glcos⁡θ\sqrt {2gl\cos \theta }2glcosθ​
  4. D
    2gl\sqrt {2gl}2gl​
View written solutionFree

Correct answer: A

AIPMT 2000 Physics - Oscillations Question 19 English Explanation


In ΔOAC, cos⁡θ=A/l\Delta OAC,\,\cos \theta = A/lΔOAC,cosθ=A/l

⇒OA=lcosθ\Rightarrow OA = lcos\theta⇒OA=lcosθ

∴\therefore∴ AB=l(1−cos⁡θ)=hAB = l\left( {1 - \cos \theta } \right) = hAB=l(1−cosθ)=h

At point, C the velocity of bob = 0. The vertical acceleration = g

∴\therefore∴ v2=2gh{v^2} = 2ghv2=2gh

⇒v=2gl(1−cos⁡θ)\Rightarrow v = \sqrt {2gl\left( {1 - \cos \theta } \right)}⇒v=2gl(1−cosθ)​

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