The string of a pendulum of length l is displ | vedclass

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(B) When the pendulum is released from $90^{\circ}$,it falls to the mean position. By the law of conservation of energy,the potential energy lost equa

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$3mg$

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(B) When the pendulum is released from $90^{\circ}$,it falls to the mean position. By the law of conservation of energy,the potential energy lost equals the kinetic energy gained:$mgl = \frac{1}{2}mv^2 \implies v^2 = 2gl$.At the mean position,the forces acting on the bob are the tension $T$ (upwards) and weight $mg$ (downwards). The net centripetal force is:$T - mg = \frac{mv^2}{l}$.Substituting $v^2 = 2gl$:$T - mg = \frac{m(2gl)}{l} = 2mg$.$T = 2mg + mg = 3mg$.Thus,the minimum strength of the string must be $3mg$.

The string of a pendulum of length $l$ is displaced through $90^{\circ}$ from the vertical and released. The minimum strength of the string required to withstand the tension as the pendulum passes through the mean position is:

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