Question

A pendulum suspended from the roof of an elevator at rest has a time period `T_(1)`, when the elevatore moves up with an acceleration `a` its time period becomes `T_(2)`, when the elevator moves down with an acceleration `a`, its time period becomes `T_(3)`, then
A. `T_(1)=(T_(2)+T_(3))/(2)`
B. `T_(1)=sqrt(T_(2)+T_(3))`
C. `T_(1)=(T_(2)+T_(3))/sqrt(T_(2)^(2)+T_(3)^(2))`
D. `T_(1)=(sqrt(2)T_(2)T_(3))/sqrt(T_(2)^(2)+T_(3)^(2))`

Answer 1

Correct Answer - D
`T_(1)=2pisqrt((l)/(g))` or `(4pi^(2)l)/(T_(1)^(2))=g` …(i)
`T_(2)=2pisqrt((l)/(g+a))(4pi^(2)l)/(T_(1)^(2))=g+a` …(ii)
`T_(3)=2pisqrt((l)/(g-a))` or `(4pi^(2)l)/(T_(3)^(2))=g-a` …(iii)
Adding (ii) and (iiii) , we have
`(4pi^(2)l)/(T_(2)^(2))+(4pi^(2)l)/(T_(3)^(2))=2g=2xx(4pi^(2)l)/(T_(1)^(2))`
On solving, `T_(1)=(sqrt(2)T_(2)T_(3))/([T_(2)^(2)+T_(3)^(2)]^(1//2))`