Question

A simple pendulum is suspended from the ceiling of a lift. when the lift is at rest its time period is t. with what acceleration should the lift be accelerated upwards in order to reduce its period to t ∕2? (g is acceleration due to gravity).

Answer 1

A simple pendulum is suspended from the ceiling of a lift , when the lift is at rest its time period is t .
Let the length of simple pendulum is L , and in rest net acceleration is g
so, time period , t = $2\pi\sqrt{\frac{L}{g}}$ -----(1)

Question said lift is now moving upward then time period of simple pendulum reduce to t/2 . Let lift is moving upward with acceleration a
Now, apply psuedo force concept ,
Then, net acceleration = (a + g)
Now, time period , t/2 = $2\pi\sqrt{\frac{L}{(a + g)}}$-----(2)

Dividing equation (2) by (1)
$\frac{t/2}{t}=\frac{2\pi\sqrt{\frac{L}{(a+g)}}}{2\pi\sqrt{\frac{L}{g}}}$
⇒1/2 = $\sqrt{\frac{g}{(a+g)}}$
squaring both sides,
⇒1/4 = g/(a + g)
⇒ a + g = 4g
⇒ a = 3g

Hence, acceleration of lift in upward direction is 3g.

Answer 2

hi there

here's your answer

plz mark as brainliest if found helpful