Question

Question #43057

Answer 1

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period `(5T)/4`

As the two pendulums start at the same time in simple harmonic motion from the mean position,their initial phase angles are zero

The bigger pendulum completes its one oscillation in time `t=(5T)/4` and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle `2pi`

If `omega` represents the angular velocity of the reference point associated with SHM of first pendulum of time period T,then the phase of the this pendulum after `(5T)/4` s will be

`omega(5T)/4=omegaT+omegaT/4`

`=2pi+(2pi)/4=2pi+pi/2`

Hence the phase difference

`2pi+pi/2-2pi=pi/2`