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(D) The total mechanical energy $E$ of a simple pendulum is the sum of its kinetic energy and potential energy, which remains constant throughout the

Vedclass · Accepted Answer

$1.41$

Vedclass · Answer

(D) The total mechanical energy $E$ of a simple pendulum is the sum of its kinetic energy and potential energy, which remains constant throughout the motion.
Given total energy $E = 0.02 	ext{ J}$.
At the equilibrium position (mean position), the potential energy of the bob is zero, so the total energy is entirely in the form of kinetic energy.
Therefore, $E = K.E_{max} = \frac{1}{2} m v_{max}^2$.
Given mass $m = 20 	ext{ g} = 0.02 	ext{ kg}$.
Substituting the values into the equation:
$0.02 = \frac{1}{2} 	imes 0.02 	imes v_{max}^2$
$1 = \frac{1}{2} v_{max}^2$
$v_{max}^2 = 2$
$v_{max} = \sqrt{2} \approx 1.414 	ext{ m/s}$.
Thus, the speed at the equilibrium position is approximately $1.41 	ext{ m/s}$.

The sum of kinetic energy and potential energy of a simple pendulum bob is $0.02 \text{ J}$. The speed of the simple pendulum bob at the equilibrium position is approximately: (Consider mass of the bob = $20 \text{ g}$) (in $\text{ m/s}$)

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