The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y=Asin\left(\pi t+?\right),$ where time is measured in second. The length of pendulum is

The equation y = acos²ωt + asin²ωt is

Two simple harmonic motion are represented by,

And 

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The ratio of their amplitudes is

For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be

The given arrangement is in vertical plane and initially the springs are in natural length. The uniform rod is hinged at end O, about which it can freely rotate. The angular frequency of small oscillations of the rod is n/8 √(k/m). Find n. (Given: m = mass of rod, L = length of rod and k = mg/L)

For an SHM oscillator, the amplitude is 5 cm and its time period is 4 seconds. The minimum time taken by the particle to pass between points which are at distances 4 cm and 3 cm from the centre of oscillation on the same side of it will be