Question

A simple pendulum in a stationary lift has time period T. What would be the effect on the time 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?

Answer 1

Time period of a simple pendulum in a stationary lift; T = 2π
L
→ (1)
g
Time period of a simple pendulum in a lift moving up with uniform acceleration is
Tup = 2π
L
→ (2)
g +a
Time period of a simple pendulum in a lift moving down with uniform acceleration is
Tdown = 2π
L
→ (3)
g −a
i) Moves up with uniform velocity, acceleration of lift at = 0
Then Top = 2π
L
= T [∵ (1)]
g
∴ No change in time period
ii) Moves down with uniform velocity a = 0
Then Tdown = 2π
L
= T [∵ (1)]
g
∴ No change in time period
Tup
(2)
⇒
=
(1)
T
g
⇒ Tup
g+a
∴ The lift moves up with uniform acceleration a, time period decreases
iv)
T
(3)
⇒ down =
(1)
T
g
⇒ Tdown > T
g −a
∴ The lift moves down with uniform acceleration a, time period increases
v) When the lift falls freely under gravity a = g
⇒ Tdown =
7.
L
=
g−g
L
= infinity