The time period of a simple pendulum when it is made to oscillate on the surface of moon

If the length of a clock pendulum increases by $0.2 \%$ due to atmospheric temperature rise, then the loss in time of clock per day is ........... $s$

What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of $10\,cm$ ($g = 9.8\, m/s^2$) ..... $m/s$

Answer the following questions:

$(a)$ Time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle:

$T=2 \pi \sqrt{\frac{m}{k}}$. A stmple pendulum executes $SHM$ approximately. Why then is the time pertodof.anondwers period of a pendulum independent of the mass of the pendulum?

$(b)$ The motion of a simple pendulum is approximately stmple harmonte for small angle oscillations. For larger angles of oscillation, a more involved analysis shows that $T$ is greater than $2 \pi \sqrt{\frac{l}{g}} .$ Think of a qualitative argument to appreciate this result.

$(c)$ A man with a wristwatch on his hand falls from the top of a tower. Does the watch give correct time during the free fall?

$(d)$ What is the frequency of oscillation of a simple pendulum mounted in a cabin that is freely failing under gravity?

A simple pendulum of length $L$ and mass (bob) $M$ is oscillating in a plane about a vertical line between angular limits $ - \varphi $ and $ + \varphi $. For an angular displacement $\theta (|\theta | < \varphi )$, the tension in the string and the velocity of the bob are $T$ and $ v$ respectively. The following relations hold good under the above conditions

A simple pendulum hanging from the ceiling of a stationary lift has a time period $T_1$. When the lift moves downward with constant velocity, the time period is $T_2$, then