Question

The length of a simple pendulum is 2 m and it is suspended from the roof of a lift. If lift is going with an acceleration of 2 m s^-2 upward. Then what will be the time period of the simple pendulum? If the rope of the lift is broken and it falls freely, then what will be the time period of the simple pendulum in this position. (g = 9.8 ms^-2)

Answer 1

l = 2 m; a = 2.0 m s^-2 (on upper side); T = ?

Lift is going upwards

R - mg = ma

or R = mg + ma = m(g + a)

= m(9.8 + 2)

= m x 11.8

If apparent acceleration is g', then Apparent weight

R = mg'

∴ mg' = m x 11.8

or g' = 11.8 m s^-2

∴ Time period of pendulum in lift

\(T = 2\pi\sqrt{\frac{l}{g}} = 2\pi\sqrt{\frac{2}{11.8}} = 2\pi\sqrt{0.16949}\) 

= 2 x 3.14 x 0.41169

or T = 2.58 s

If the rope of the lift is broken, then lift falls freely

∴ pendulum will be in position of weightlessness, or g' = 0

\(\therefore\ T = 2\pi\sqrt{\frac{2}{0}} = \infty\) (infinite)

Based on spring-block system