The bob of a pendulum is released from a horizontal position. If the lenght of the pendulum is 1.5 m what is the speed with which the bob arrives at the lowermost point, given that it dissipated `5%` of its initial energy against air resistance?

Explanation

Decrease in RE. of the bob in falling through a height `l` = Increase in its K.E.
`mgl = 1/2 mv^(2)`
`mv^(2) = 2 mgl`………..(1)
`v = sqrt(2 gl)`

At the lowest position B, the forces acting on the bob are the tension in the string 7/acting vertically upward and the weight of the bob acting vertically downward. The resultant force gives the centripetal force.
`(mv^(2))/l = (T-mg)`, because the radius of the circular path r=L
`T = mg + (mv^(2))/l`. But from equation (i) `mv^(2) = mgl`
`T = mg + (2mgl)/l = 3mg`