Show that the time period of a simplw pendulam of infinite length is given by `t=2pi sqrt((R)/(g))` where R is the radius of the earth

The time period of a simple pendulum of infinite length is (R=radius of earth).

Show that the period of a simple pendium is given by T=2pi sqrt((m)/(m),(1)/(g)) where m is the 'internial mass' and m is the 'gravitional mass' of the bob.

The ratio of the time period of a simple pendulum of length l_(0) with a pendulum of infinite length is : (Where, R is the radius of earth)

Show that the expression of the time period T of a simple pendulum of length l given by T = 2pi sqrt((l)/(g)) is dimensionally currect

The time period of a simple pendulum is given by T = 2pi sqrt((l)/(g)) where 'l' is the length of the pendulum and 'g' is the acceleration due to gravity at that place. (a) Express 'g' in terms of T (b) What is the length of the seconds pendulum ? (Take g = 10 m s^(-2) )

The time t of a complete oscillation of a simple pendulum of length l is given by the equation T=2 pi sqrt((1)/(g)) where g is constant.What is the percentage error in T when l is increased by 1%?

A statellite is launched into a circular orbit, (R )/(4) above the surface of the earth. The time period of revolution is T=2pi(n)^(3//2)sqrt((R )/(g)) . Where R is the radius of the earth and g is the acceleration due to gravity. Then what is the value on n?

Show that the time period of a compound pendulam is minimum when it is suspended from a point at a distance equal to its radius of gyration.

The time period of a simple pendulum is given by the formula, T = 2pi sqrt(l//g) , where T = time period, l = length of pendulum and g = acceleration due to gravity. If the length of the pendulum is decreased to 1/4 of its initial value, then what happens to its frequency of oscillations ?