Question #43057

1 Answer
Mar 29, 2017

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