Calculate the kinetic energy of bob of a simple pendulum when it reaches the mean position after releasing from the extreme position as shown in the figure.

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Solution

$Δ K = - Δ U$
$= m g Δ h$
$= 0.1 \times 10 \times 0.1 (1 - c o s 60^{\circ})$
$= 0.1 \times 10 \times 0.1 \times \frac{1}{2}$
$= 0.05 J$

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Calculate the kinetic energy of bob of a simple pendulum when it reaches the mean position after releasing from the extreme position as shown in the figure.

ΔK=−ΔU =mgΔh =0.1×10×0.1(1−cos 60∘) =0.1×10×0.1×12 =0.05 J

$Δ K = - Δ U$
$= m g Δ h$
$= 0.1 \times 10 \times 0.1 (1 - c o s 60^{\circ})$
$= 0.1 \times 10 \times 0.1 \times \frac{1}{2}$
$= 0.05 J$