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

If the length of an oscillating simple pendulum is made $\frac{1}{3}$ times at a place while keeping the amplitude the same,then its total energy $(E)$ will be: (in $E$)

$A$ simple pendulum with a bob of mass $40 \ g$ and charge $+2 \ \mu C$ makes $20$ oscillations in $44 \ s$. $A$ vertical electric field of magnitude $4.2 \times 10^4 \ NC^{-1}$ pointing downward is applied. The time taken by the pendulum to make $15$ oscillations in the electric field is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $s$)

Time period of a simple pendulum is $T$ inside a lift when the lift is stationary. If the lift moves upwards with an acceleration $g / 2$,the time period of the pendulum will be

The bob of a simple pendulum executes simple harmonic motion in water with a period $t$,while the period of oscillation of the bob is ${t_0}$ in air. Neglecting the frictional force of water and given that the density of the bob is $(4/3) \times 1000 \ kg/m^3$. What relationship between $t$ and ${t_0}$ is true?

Two simple pendulums of lengths $1.44 \, m$ and $1 \, m$ start swinging together. After how many vibrations will they again start swinging together?