Simple Harmonic motion

Use equation: F - mg = ma
where F is the tension on the pendulumn suspension
m is the mas of the pendulumn bob
g is the acceleration due to gravity
a is the acceleration of the lift

When the lift is moving downward with a constant speed, a = 0 m/s2
hence, F = mg

The pendulum oscillates as if it was at rest, hence T1 = T0 = 2.pi.square-root[L/g], where L is the pendulum length

When the lift accelerates downward,
mg -'F = ma
i.e. F' = m(g - a)
hence, the tension is smaller than if the lift is at rest,
the apparent acceleration due to gravity g' is
F' = mg' = m(g-a)
i.e. g' = (g - a)

Hence, T2 = 2.pi.square-root[L/(g-a)]
Clearly, T2 > T0 = T1

When the lift descends at a steady speed, the apparent acceleration in the lift is still g.

Thus T0 = T1

When the lift descends at a constant downward acc., the apparent acceleration in the lift is smaller than g.

Thus By the formula T = 2π√(l/g')

T2 > T0

Ans: A