The bob of a simple pendulum having length ' ℓ' is displaced from its equilibrium position by an

Name: Problem 1344, Chapter 10: JEE Physics
Uploaded: 2021-10-02T05:28:30-08:00
Duration: 2 min 23 s
Channel: Supratim Pal
Description: The bob of a simple pendulum having length ' ℓ' is displaced from its equilibrium position by an angle of θand released. If the velocity of the bob, while passing through its equilibrium position is v, then v=………(A) √({2 g ℓ(1-cosθ)}) (B) √({2 g ℓ(1+sinθ)}) (C) √({2 g ℓ(1-sinθ)}) (D) √({2 g ℓ(1+cosθ)})

step 1 Hello and welcome to this video solution of numerate. So it is given that the bob of a simple pe

Question

The bob of a simple pendulum having length ' $\ell$ ' is displaced from its equilibrium position by an angle of $\theta$ and released. If the velocity of the bob, while passing through its equilibrium position is $\mathrm{v}$, then $\mathrm{v}=\ldots \ldots \ldots$ (A) $\sqrt{\{2 g \ell(1-\cos \theta)\}}$ (B) $\sqrt{\{2 g \ell(1+\sin \theta)\}}$ (C) $\sqrt{\{2 g \ell(1-\sin \theta)\}}$ (D) $\sqrt{\{2 g \ell(1+\cos \theta)\}}$

   The bob of a simple pendulum having length ' $\ell$ ' is displaced from its equilibrium position by an angle of $\theta$ and released. If the velocity of the bob, while passing through its equilibrium position is $\mathrm{v}$, then $\mathrm{v}=\ldots \ldots \ldots$
(A) $\sqrt{\{2 g \ell(1-\cos \theta)\}}$
(B) $\sqrt{\{2 g \ell(1+\sin \theta)\}}$
(C) $\sqrt{\{2 g \ell(1-\sin \theta)\}}$
(D) $\sqrt{\{2 g \ell(1+\cos \theta)\}}$

JEE Physics

StemEZ 1st Edition

Chapter 10, Problem 1344

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Solved by Expert Supratim Pal

10/02/2021

Step 1

The bob is then released. We need to find the velocity of the bob when it passes through its equilibrium position. Show more…

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The bob of a simple pendulum having length ' $\ell$ ' is displaced from its equilibrium position by an angle of $\theta$ and released. If the velocity of the bob, while passing through its equilibrium position is $\mathrm{v}$, then $\mathrm{v}=\ldots \ldots \ldots$ (A) $\sqrt{\{2 g \ell(1-\cos \theta)\}}$ (B) $\sqrt{\{2 g \ell(1+\sin \theta)\}}$ (C) $\sqrt{\{2 g \ell(1-\sin \theta)\}}$ (D) $\sqrt{\{2 g \ell(1+\cos \theta)\}}$

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Key Concepts

Conservation of Energy

In physics, the conservation of energy principle states that energy cannot be created or destroyed but only transformed from one form to another. For a simple pendulum, the gravitational potential energy at its highest point is completely converted into kinetic energy as it swings through the equilibrium position. This conversion is key to determining the velocity of the pendulum bob at different points in its trajectory.

Gravitational Potential Energy in a Pendulum

The gravitational potential energy of a pendulum bob is related to its height above the lowest point, which, in the case of a pendulum, can be determined using the cosine of the displacement angle. The drop in height from the displaced position to the equilibrium position is derived from the pendulum length and the cosine of the release angle, which is then used in the energy conservation equation to calculate the velocity at the lowest point.

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