If the length of a simple pendulum is increased by x, then its time period increases from `T_(1)` to `T_(2)`.
Prove that `g=(4pi^(2)x)/(T_(2)^(2)-T_(1)^(2))`.

The time period of a simple pendulum is T. When the length is increased by 10 cm, its period is T_(1) . When the length is decreased by 10 cm, its period is T_(2) . Prove that the relation between T,T_(1)andT_(2) is T_(1)^(2)+T_(2)^(2)=2T^(2) .

Let f and g be periodic functions with the periods T_(1) and T_(2) respectively. Then f+g is

A particle at the end of a spring executes simple harmonic motion with a period t_(1) , while the corresponding period for another spring is t_(2) . If the period of oscillation with the two springs in series is T, then prove that t_(1)^(2)+t_(2)^(2)=T^(2)

If the time taken by a particle moving at constant acceleration to travel equal distances along a straight line are t_(1), t_(2), t_(3) respectively prove that (1)/(t_(1))-(1)/(t_(2))+(1)/(t_(3))=(3)/(t_(1)+t_(2)+t_(3)) .

If the time period of a simple pendulum of effective length L is T, then the effective length of a simple pendulum having time period 2T will be