Question

A simple pendulum of length l is suspended through the ceiling of an elevator. Find the time period of small oscillations if the elevator (a) is going up with an acceleration a₀ (b) is going down with an acceleration a₀ and (c) is moving with a uniform velocity.

Answer 1

Explanation ⇒  The Formula of the time period of S.H.M. of pendulum when it is making small oscillations is given by,

 $T = 2\pi \sqrt{\frac{l}{g'} }$

g' is the effective acceleration here.

(a). When the elevator is moving up with an acceleration a₀, then g effective will given by the sum of g and a₀.

g' = g + a

∴  $T = 2\pi \sqrt{\frac{l}{g + a} }$

(b). When the elevator is moving down with an acceleration a₀, then g effective will given by the difference of g and a₀.

g' = g - a

$T = 2\pi \sqrt{\frac{l}{g - a} }$

(c). When the elevator is moving down with an uniform velocity, then net acceleration will be zero. Thus, effective acceleration will be equal to g only.

$T = 2\pi \sqrt{\frac{l}{g} }$

Hope it helps.