A simple pendulum is oscillating in a stationery lift. When the lift falls freely, the frequency of oscillations of the pendulum is

Options

A. zero
B. `upsilon`
C. `2upsilon`
D. infinite

Answer

A

Explanation

When the lift is at rest, the frequency of oscillation of simple pendulum is
`upsilon=(1)/(2pi)sqrt((g)/(L)) `
When the lift is at rest, the frequency of oscillation of simple pendulum is
`upsilon(1)/(2pi)sqrt((g)/(L))`
When the lift falls freely, then effective value of acceleration due to gravity is
`g'=g-a=g-g=0`
`thereforeupsilon=2pisqrt((0)/(L))=0`