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(B) Let the mass of the pendulum bob be $m$ and the length of the string be $l$. The center of mass of the string is at $l/2$ from the pivot.When the

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$3 m g$

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(B) Let the mass of the pendulum bob be $m$ and the length of the string be $l$. The center of mass of the string is at $l/2$ from the pivot.When the pendulum is displaced by $90^{\circ}$ and released,the change in potential energy is converted into kinetic energy.Using the law of conservation of energy: $m g (l/2) = (1/2) m v^2$.This gives $v^2 = g l$.At the mean position,the tension $T$ in the string must balance the weight $m g$ and provide the necessary centripetal force $F_c = m v^2 / r$,where $r = l/2$ is the distance of the center of mass from the pivot.Therefore,$T = m g + (m v^2) / (l/2)$.Substituting $v^2 = g l$ into the equation:$T = m g + (2 m / l) \cdot (g l) = m g + 2 m g = 3 m g$.Thus,the minimum strength of the string must be $3 m g$.

$A$ string of a pendulum of length $l$ is displaced through $90^{\circ}$ from its vertical and released. Then,the minimum strength of the string needed to withstand the tension as the pendulum passes through its mean position is

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