The period of a simple pendulum measured inside a stationary lift is found to be $T$. If the lift starts accelerating upwards with acceleration of $g/3,$ then the time period of the pendulum is

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 ……….. $s$

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