Physics, asked by danishsoni5804, 10 months ago


A pendulum bob on a 2 m string is displaced 60° from the
vertical and then released. What is the speed of the bob as
it passes through the lowest point in its path (JIPMER 1999)​

Answers

Answered by IamIronMan0
0

Answer:

Use mechanical energy conservation

mgh =  \frac{1}{2} m {v}^{2}  \\  \\  {v}^{2}  = 2g(l - l\cos(60))  \\   \\ \cos(60)  =  \frac{1}{2}    \\ \\  {v}^{2} = 2g(l -  \frac{l}{2} ) = gl \\  \\ v =  \sqrt{gl}  =  \sqrt{  10 \times 2}  = 2 \sqrt{5}  \:  \frac{m}{s}

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