Two pendulums begin to swing simultaneously. If the ratio of the frequency of oscillations of the two is $7 : 8$,then the ratio of lengths of the two pendulums will be

$A$ coin is placed on a horizontal plate. The plate performs $S.H.M.$ vertically with an angular frequency $\omega$. The amplitude $A$ of oscillations is gradually increased. The coin will lose contact with the plate for the first time when the amplitude is ($g =$ acceleration due to gravity).

Two pendulums of length $121 \ cm$ and $100 \ cm$ start vibrating in phase. At some instant,the two are at their mean position in the same phase. The minimum number of vibrations of the shorter pendulum after which the two are again in phase at the mean position is:

$A$ simple pendulum has a time period $T_1$. The point of suspension is now moved upward according to the equation $y = kt^2$,where $k = 1 \ m/s^2$. If the new time period is $T_2$,then the ratio $\frac{T_1^2}{T_2^2}$ will be (take $g = 10 \ m/s^2$):

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?

$A$ 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).