question_answer

A simple pendulum consisting of an inextensible length \[l\] and mass m is oscillating in a stationary lift. The lift then accelerates upwards with a constant acceleration of \[\mathbf{4}.\mathbf{5}\text{ }\mathbf{m}/{{\mathbf{s}}^{\mathbf{2}}}\]. Write expression for the time period of simple pendulum in two cases. Does the time period increase, decrease or remain the same, when lift is accelerated upwards?

Answer:

                When the lift is stationary \[T=2\pi \sqrt{\frac{l}{g}}\]  (ii) When the lift accelerates upwards with an acceleration of\[\text{4}.\text{5 m}/{{\text{s}}^{\text{2}}}\], \[T'=2\pi \sqrt{\frac{l}{g+a}}=2\pi \sqrt{\frac{l}{g+4.5}}\] Clearly, the time period decreases when the lift accelerates upwards.