The velocity of a simple pendulum is maximum at

The time period of a simple pendulum in a stationary lift is $T$. If the lift accelerates with $\frac{g}{6}$ vertically upwards,then the new time period will be:
(where $g =$ acceleration due to gravity)

The bob of a swinging seconds pendulum (one whose time period is $2 \,s$) has a small speed $v_{0}$ at its lowest point. Its height from this lowest point $2.25 \,s$ after passing through it is given by

$A$ simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration $a$. The time period is given by $T = 2\pi \sqrt{\frac{l}{g'}}$,where $g'$ is equal to:

The path length of oscillation of a simple pendulum of length $1 \,m$ is $16 \,cm$. Its maximum velocity is (Take $g = \pi^2 \,m/s^2$).

$A$ man measures the period of a simple pendulum inside a stationary lift and finds it to be $T$ seconds. If the lift accelerates upwards with an acceleration of $\frac{g}{4}$,then the period of the pendulum will be