The velocity of a simple pendulum is maximum at

$A$ pendulum clock is set to give correct time at sea level. This clock is moved to a hill station at an altitude of $2500 \ m$ above sea level. In order to keep correct time at the hill station,the length of the pendulum:

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its time period is $T$. With what acceleration should the lift be accelerated upwards in order to reduce its period to $T/2$ (in $g$)? ($g$ is acceleration due to gravity).

The acceleration due to gravity on the surface of the moon is $1.7 \; m s^{-2}$. What is the time period of a simple pendulum on the surface of the moon if its time period (in $s$) on the surface of the earth is $3.5 \; s$? ($g$ on the surface of the earth is $9.8 \; m s^{-2}$)

$A$ simple pendulum of length $1\,m$ is allowed to oscillate with an amplitude of $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be: (use $g = \pi^2$)

In a stationary lift,the time period of a simple pendulum is $T$. If the lift starts accelerating downwards with an acceleration of $\frac{g}{4}$,then the new time period of the pendulum will be: