The time period of a simple pendulum inside a stationary lift is √3s. When the lift moves downward with an acceleration g4, find the time period of simple pendulum.
A
1s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
3s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B2s Time period of pendulum when lift is stationary, T=2π√lg=√3s
For the lift accelerating downwards, a=g4,
Effective value of acceleration due to gravity is, ⇒g′=g−a=g−g4=3g4
⇒Time period of pendulum in accelerating lift will be: T′=2π√lg′ T′=2π√4l3g=√43×2π√lg T′=√43×√3=2s