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) \(\sqrt{\\{2 g \ell(1+\cos \theta)\\}}\)

Short answer

The short answer is: \(v = \sqrt{2gl(1-\cos \theta)}\).

Step-by-step solution

STEP 1: Identify the initial and final points

In this situation, we will consider two points: - Initial point: when the pendulum is released from an angle of θ with respect to vertical. - Final point: when the pendulum passes through its equilibrium position (vertical).

STEP 2: Calculate the initial potential and kinetic energy

Initially, the bob is at an angle θ and at rest (with an initial velocity of 0). So, the initial kinetic energy is zero. We need to find the initial potential energy. As the bob is raised by a vertical height of \((l-l\cos \theta)\), the initial potential energy is: \[PE_{i}=mgh=mgl(1-\cos \theta)\]

STEP 3: Calculate the final potential and kinetic energy

As the pendulum passes through its equilibrium position, its gravitational potential energy becomes zero. We need to find the final kinetic energy, which is given by: \[ KE_{f} = \frac{1}{2}mv^2\]

STEP 4: Use the conservation of energy principle

According to the conservation of mechanical energy: \[ PE_{i} + KE_{i} = PE_{f} + KE_{f}\] Since \(PE_{i} = mgl(1-\cos \theta)\), \(KE_{i} = 0\), \(PE_{f} = 0\), and \(KE_{f} = \frac{1}{2}mv^2\), the equation becomes: \[mgl(1-\cos \theta) = \frac{1}{2}mv^2\]

STEP 5: Solve the equation and find the velocity

Now, we need to solve the equation for the velocity 'v': \[v^2 = 2gl(1-\cos \theta)\] Taking the square root of both sides, we get: \[v = \sqrt{2gl(1-\cos \theta)}\] Comparing the expression for velocity with the given options, we can conclude that the correct answer is: (A) \(v = \sqrt{\\{2 g \ell(1-\cos \theta)\\}}\)