The mass of the bob of a simple pendulum of l | vedclass

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(A) Let the horizontal position be $A$ and the lowest position be $B$. The length of the pendulum is $L$.By the law of conservation of energy,the pote

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$\sqrt{2gL}$ and $3mg$

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(A) Let the horizontal position be $A$ and the lowest position be $B$. The length of the pendulum is $L$.By the law of conservation of energy,the potential energy at $A$ is equal to the kinetic energy at $B$:$m g L = \frac{1}{2} m v^2$$v^2 = 2 g L$$v = \sqrt{2 g L}$At the lowest position $B$,the forces acting on the $bob$ are the tension $T$ (upwards) and the weight $mg$ (downwards). The net force provides the necessary centripetal force:$T - m g = \frac{m v^2}{L}$$T = m g + \frac{m (2 g L)}{L}$$T = m g + 2 m g = 3 m g$Thus,the speed is $\sqrt{2 g L}$ and the tension is $3 m g$.

The mass of the $bob$ of a simple pendulum of length $L$ is $m$. If the $bob$ is released from its horizontal position,then the speed of the $bob$ and the tension in the thread at the lowest position of the $bob$ will be respectively:

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