A simple pendulum in a stationary lift has time period T. What would be the effect on the period when the lift (i) moves up with uniform velocity (ii) moves down with uniform velocity (iii) moves up with uniform acceleration 'a' (iv) moves down with uniform acceleration 'a' (v) begins to fall freely under gravity?

Explanation

i) When the lift moves up with uniform velocity i.e. `a=0`. There would be no change in the time period of a simple pendulum.
ii) When the lift moves down with uniform velocity i.e. `a=0`, there would be no change in the time period of a simple pendulum.
iii) When lift is moving up with acceleration 'a' then relative acceleration `=g+a`
`therefore` Time period, `T=2pisqrt((l)/(g+a))` so when lift is moving up with uniform acceleration time period of pendulum in it decreases.
iv) When lift is moving down with acceleration 'a' time period, `T=2pisqrt((l)/(g-a))` (`g-a=` relative acceleration of pendulum)
So time period of pendulum in the lift decreases.
v) If the falls freely, `a=g` then the time period of a simple pendulum becomes infinite.