A simple pendulum is placed inside a lift, which is moving with a uniform acceleration. If the time periods of the pendulum while the lift is moving upwards and downwards are in the ratio 1 : 2, then the acceleration of the lift is (Acceleration due to gravity, g = 10 `ms^(-2)` )

Options

A. `6 ms^(-2)`
B. 0 `ms^(-2)`
C. 3`ms^(-2)`
D. 2 `ms^(-2)`

Answer

A

Explanation

Given,
time period of the pendulum while the lift is moving upwards and downward are in the ratio,
` T_(1) : T_(2) = 1 : 2 `
Acceleration due to gravity, g = 10 `m//s^(2)`
We know that,
If the lift is moving upward , then total time-period.
` T_(1) = 2pi sqrt( (l)/(g + a)) " " ` ..... (i)
when the lift is moving downwards, then the total time period,
`T_(2) = 2pi sqrt((l)/(g - a)) " " ` .... (ii)
By dividing Eq. (i) to (ii) , we get
`therefore" " (T_(1))/(T_(2)) = sqrt( (g - a)/(g + a))`
now, `" " (1)/(2) = sqrt((g - a)/(g + a))`
square on the both sides, we get
or ` ((1)/(2))^(2) = (g -a)/(g + a) or (g - a)/(g + a) = (1)/(4) `
or `" " 4g - 4a = g + a or 3g = 5 a `
or `" " a = (3g)/(5)`
`rArr " " a = (30)/(5) = 6 m//s^(2)`
So, the acceleration of the lift is 6 `m//s^(2)`