A second pendulum is taken in a carriage. Find the period of oscillation when the carriage moves with an acceleration of `4*2ms^(-2)` (i) vertically upwards (ii) vertically downwards and (ii) in a horizontal direction.

Explanation

(a) Time period of pendulum is
`T = 2pisqrt((l)/(g))`
In case of second.s pendulum, T = 2 s,br> `rArr 2 = 2pi sqrt((l)/(g))`
`rArr 1 = pi^(2) (l)/(g)`
`rArr l = (g)/(pi^(2)) = (9.8)/(pi^(2))`
(i) When the elevator moves downwards with an acceleration `a = 3 m//s^(2)`, the time period is
`T = 2pisqrt((l)/(g-a)) = 2pisqrt((9.8)/(pi^(2)(9.8-3)))`
` = (2pi)/(pi) sqrt((9.8)/(6.8)) = 2 xx 1.2 = 2.4s`
(ii) When elevator moves upwards, then
`T . = 2pisqrt((l)/(g+a)) = 2pisqrt((9.8)/(pi^(2)(9.8 + 3))`
` = (2pi)/(pi) sqrt((9.8)/(12.8)) = 2 xx 0.87 = 1.75`
(b) When the pendulum is taken in a carriage moving horizontal with a = 3 `m//s^(2)` a and g will be aqt right angle to each other. So, the net acceleration is
`a = sqrt((g^(2) + a^(2)) = sqrt(9.8 _ 3)`
` = sqrt(96.04 + 9) = sqrt(105.04) = 10.25 m//s^(2)`
`therefore` Time period, T . = `2pi sqrt((l)/(a)) = 2pisqrt((9.8)/(pi^(2) xx 10.25))`
` = 2 xx 0.9778 = 1.96` s