If a simple pendulum is taken to a place where $g$ decreases by $4\%$,then the time period

$A$ pendulum clock is running fast. To correct its time,we should

$A$ simple pendulum,suspended from the ceiling of a stationary van,has time period $T$. If the van starts moving with a uniform velocity,the period of the pendulum will be

If the length of a seconds pendulum becomes $\frac{1}{3}$ of its original length,what will be its new periodic time?

The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length $l$. If both bobs are charged with a charge $q$ each, what is the time period of the pendulum? (Ignore the radii of the bobs.)

Answer the following questions:
$(a)$ The time period of a particle in $SHM$ depends on the force constant $k$ and mass $m$ of the particle: $T=2 \pi \sqrt{\frac{m}{k}}$. $A$ simple pendulum executes $SHM$ approximately. Why then is the time period of a pendulum independent of the mass of the pendulum?
$(b)$ The motion of a simple pendulum is approximately simple harmonic 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 falling under gravity?