# Question #43057

Mar 29, 2017

Assuming bigger pendulum as the pendulum of bigger time period i.e.having time period $\frac{5 T}{4}$

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 $t = \frac{5 T}{4}$ and comes back to the mean position.During this time the point on reference circle associated with SHM rotates by an angle $2 \pi$

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 $\frac{5 T}{4}$ s will be

$\omega \frac{5 T}{4} = \omega T + \omega \frac{T}{4}$

$= 2 \pi + \frac{2 \pi}{4} = 2 \pi + \frac{\pi}{2}$

Hence the phase difference

$2 \pi + \frac{\pi}{2} - 2 \pi = \frac{\pi}{2}$