The bob of a pendulum of length ' l ' is pulled aside from its equilibrium position through an angle ' θ ' and then released. The bob will then pass through its equilibrium positio

Question

The bob of a pendulum of length ' l ' is pulled aside from its equilibrium position through an angle ' θ ' and then released. The bob will then pass through its equilibrium position with speed ' v ', where ' v ' equal to ( g = acceleration due to gravity)

TestHub · Accepted Answer

Let the equilibrium position be the reference level for potential energy.

When the pendulum of length $l$ is pulled aside by an angle $θ$ , the vertical height $h$ of the bob above the equilibrium position is given by:

$h = l - l cos θ = l (1 - cos θ)$

At the release point, the kinetic energy is zero and the potential energy is $m g h$ .

At the equilibrium position, the potential energy is zero and the kinetic energy is $\frac{1}{2} m v^{2}$ .

By the principle of conservation of mechanical energy:

$\frac{1}{2} m v^{2} = m g h$

$\frac{1}{2} m v^{2} = m g l (1 - cos θ)$

$v^{2} = 2 g l (1 - cos θ)$

$v = 2 g l (1 - cos θ)$

Answer: $2 g l (1 - cos θ)$

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