inside a stationary lift the time period of a simple pendulum is 4 sec what will be the time period of the pendulum, if the lift (1) moves upward with an acceleration of 4m /sec ^2 (2) moves downward with an acceleration of 4m /sec ^2.

Time period, T = 2π sqrt(l/g)Time period of the pendulum when the lift moves upward with an acceleration ‘a’ = 2π sqrt(l/(g+a))We have, T = 4sec = 2π sqrt(l/10)  ------------------(1)= 2π sqrt(l/(10+4))  -----------(2)^’ , T²When acceleration upward with a= 4m/sDivide the equation (2) by (1) / 4] = [sqrt (10/14)]^’[T  = 4* [sqrt (10/14)] = 3.38 sec² [1] Therefore, time period of the pendulum when the lift moves upward with an acceleration of 4m/sTime period of the pendulum when the lift moves downward with an acceleration ‘a’ = 2π sqrt(l/(g-a))  = 4* [sqrt (10/6)] = 5.16 sec² Proceeding as mentioned above, we get: [2] Time period of the pendulum when the lift moves upward with an acceleration of 4m/s

  = 4* [sqrt (10/14)] = 3.38 sec.................² Time period of the pendulum when the lift moves upward with an acceleration of 4m/s  = 4* [sqrt (10/6)] = 5.16 sec.....² Time period of the pendulum when the lift moves upward with an acceleration of 4m/s