Question

State the law of conservation of energy. Show that the energy of a viberating pendulum is conserved

Answer 1

At point ‘A’ velocity of the bob of simple pendulum is zero. Therefore, K.E. at point ‘A’ = 0. Since the bob is at a height (h), Therefore, P.E. of the bob will be maximum. i.e.
P.E. = mgh.
Energy total = K.E. + P.E
Energy total = 0 + mgh
Energy total = mgh 
This shows that at point A total energy is potential energy.

 

If we release the bob of pendulum from point ‘A’, velocity of bob gradually increases, but the height of bob will decreases from point to the point. At point ‘M’ velocity will become maximum and the height will be nearly equal to zero.
Thus , 
K.E. = maximum = 1/2mV2 but P.E. = 0. 
Energy total = K.E. + P.E
Energy total = 1/2mV2 + 0
Energy total = 1/2mV2 
This shows that the P.E. at point is completely converted into K.E. at point ‘M’. 

At point M the bob of Pendulum will not stop but due to inertia, the bob will moves toward the point ‘B’. As the bob moves from ‘M’ to ‘B’, its velocity gradually decreases but the height increases. At point ‘B’ velocity of the bob will become zero.
Thus K.E. at point ‘B’ = 0 but P.E. = max.
P.E. = mgh.
Energy total = K.E. + P.E. 
Energy total = 0 + mgh
Energy total = mgh
This shows that at point B total energy is again potential energy.

Above analysis indicates that the total energy during the motion does not change. I.e. the motion of the bob of simple pendulum is according to the law of conservation of energy