In $S.H.M.$,a simple pendulum oscillates with frequency $f$. If the length of the pendulum is increased by three times its original length,then the frequency of oscillation of the pendulum will be

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\alpha$,is given by

$A$ simple pendulum of length $l$ and mass $m$ of the bob is suspended in a car that is travelling with a constant speed $v$ around a circular path of radius $R$. If the pendulum undergoes oscillations with small amplitude about its equilibrium position,the frequency of its oscillations will be

Given below are two statements:
Statement $I :$ $A$ second's pendulum has a time period of $1$ second.
Statement $II :$ It takes precisely one second to move between the two extreme positions.
In the light of the above statements,choose the correct answer from the options given below:

The bob of a simple pendulum having length $l$ is displaced from the mean position to an angular position $\theta$ with respect to the vertical. If it is released,then the velocity of the bob at the equilibrium position is:

Two masses $M_{A}$ and $M_{B}$ are hung from two strings of length $l_{A}$ and $l_{B}$ respectively. They are executing $SHM$ with frequency relation $f_{A}=2 f_{B}$,then the relation is: