The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

If a simple pendulum is released from the given position,find the velocity of the bob when it reaches the lowest position. (Take $g = 10 \ m/s^2$) (in $m/s$)

$A$ simple pendulum oscillating in air has period $T.$ The bob of the pendulum is completely immersed in a non-viscous liquid. The density of the liquid is $\frac{1}{16}$ of the material of the bob. If the bob is inside the liquid all the time,its period of oscillation in this liquid is:

The bob of a simple pendulum of length $L$ has a mass $m$ and charge $q$. The pendulum is suspended between the plates of a charged parallel plate capacitor. The direction of the electric field is shown in the figure. The period of oscillations of the simple pendulum is (acceleration due to gravity $g > qE/m$):

$A$ 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?

$A$ piece of wood has length,breadth and height,'$a$','$b$' and '$c$' respectively. Its relative density is '$d$'. It is floating in water such that the side '$a$' is vertical. It is pushed down a little and released. The time period of $S.H.M.$ executed by it is ($g=$ acceleration due to gravity).