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. $ \sqrt{ 2 gl ( 1 - \cos \, \theta)} $
• B. $ \sqrt{ 2 gl ( 1 + \cos \, \theta)} $
• C. $ \sqrt{ 2 gl \cos \, \theta} $
• D. $ \sqrt { 2gl}$

Answer 1

The Correct Option is A

The question involves calculating the velocity of the bob of a simple pendulum at its equilibrium position. Let's solve this step-by-step.

1. Understand the concept of mechanical energy conservation in the context of a simple pendulum. A pendulum in motion possesses both potential and kinetic energy. At the highest point, all of its energy is potential, and at the equilibrium (mean) position, it is entirely kinetic.
2. Initially, when the pendulum is displaced to an angular position \theta, the height of the bob above the equilibrium position is given by: h = l(1 - \cos \theta).
3. The potential energy at this angular position is: PE = mgh = mg \cdot l(1 - \cos \theta), where m is the mass of the bob and g is the acceleration due to gravity.
4. On releasing, the bob swings down to the equilibrium position. At this point, potential energy is converted entirely into kinetic energy. Thus, we equate the potential energy to kinetic energy: KE = \frac{1}{2}mv^2 = mg \cdot l(1 - \cos \theta).
5. Solving for the velocity v: \frac{1}{2}mv^2 = mg \cdot l(1 - \cos \theta)
   • Cancel the mass m from both sides: \frac{1}{2}v^2 = gl(1 - \cos \theta).
   • Further simplify to solve for v: v^2 = 2gl(1 - \cos \theta).
   • Finally, take the square root of both sides to find v: v = \sqrt{2gl(1 - \cos \theta)}.

Thus, the velocity of the bob at the equilibrium position is \sqrt{2gl(1 - \cos \theta)}.

Therefore, the correct answer is:

\sqrt{ 2 gl ( 1 - \cos \, \theta)}