Question

Nergy of shm 9. A simple pendulum performs s.H.M. About x -owith an amplitude a and time period t ofthe pendulum at x=a/2will be- 2 ta 3 2t 3r a

Answer 1

Therefore, d2x/dt2 +ω2 x = 0

Hence, acceleration of S.H.M. = d2x/dt2 = – ω2 x (I)