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

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(B) Let the mass of the bob be $m$ and the length of the string be $l$. When the pendulum is displaced by $90^o$ and released,the potential energy at

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$3\,mg$

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(B) Let the mass of the bob be $m$ and the length of the string be $l$. When the pendulum is displaced by $90^o$ and released,the potential energy at the initial position is converted into kinetic energy at the mean position. By the law of conservation of energy: $mgl = \frac{1}{2}mv^2$,which gives $v^2 = 2gl$. 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 $v^2 = 2gl$,we get $T - mg = \frac{m(2gl)}{l} = 2mg$. Therefore,$T = 2mg + mg = 3mg$. Thus,the minimum strength of the string must be $3mg$.

The string of a pendulum of length $l$ is displaced through $90^o$ from the vertical and released. Then,the minimum strength of the string in order to withstand the tension as the pendulum passes through the mean position is

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