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

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(A) Let the horizontal position be the reference level for potential energy $(PE = 0)$.At the horizontal position,the total energy is $E_i = PE + KE =

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

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(A) Let the horizontal position be the reference level for potential energy $(PE = 0)$.At the horizontal position,the total energy is $E_i = PE + KE = 0 + 0 = 0$.At the lowest position,the bob is at a vertical distance $L$ below the reference level,so its potential energy is $PE_f = -mgL$.Let the speed at the lowest position be $v$. The kinetic energy is $KE_f = \frac{1}{2}mv^2$.By the law of conservation of energy,$E_i = E_f$:$0 = -mgL + \frac{1}{2}mv^2$$mgL = \frac{1}{2}mv^2$$v^2 = 2gL \Rightarrow v = \sqrt{2gL}$.At the lowest 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 speed is $\sqrt{2gL}$ and the tension is $3mg$.

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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