Two simple pendulums of length $1\, m$ and $4\, m$ respectively are both given small displacement in the same direction at the same instant. They will be again in phase after the shorter pendulum has completed a number of oscillations equal to

Two pendulums differ in lengths by $22 \ cm$. They oscillate at the same place such that one of them makes $15 \ oscillations$ and the other makes $18 \ oscillations$ during the same time. The lengths (in $cm$) of the pendulums are

$A$ pendulum suspended from the ceiling of a train has a period $T$,when the train is at rest. When the train is accelerating with a uniform acceleration $a$,the period of oscillation will

$A$ simple pendulum is taken to a place where its distance from the Earth's surface is equal to the radius of the Earth. Calculate the time period of small oscillations if the length of the string is $4.0 \ m$. (Take $g = \pi^2 \ m/s^2$ at the surface of the Earth.) (in $s$)

$A$ positively charged pendulum is oscillating in a uniform electric field pointing upwards. How does its time period compare to the time period when it oscillates without an electric field?

$A$ simple pendulum of length $L$ has mass $m$ and it oscillates freely with amplitude $A$. At the extreme position,its potential energy is ($g =$ acceleration due to gravity).