Question

Two pendulums of lengths 100cm and 110.25cm start oscillatting in phase simultaneously.After how many oscillation will they again be in phase together? how?

Answer 1

The pendulums initially started off in phase. But since their time periods are different one pendulum (with smaller time period) will start gaining over the other with larger time period.

L1 = 100cm
L2 = 110.25 cm

\[T _{1}=2\pi \sqrt{L _{1}/g}\]
\[T _{2}=2\pi \sqrt{L _{2}/g}\]

T1>T2 so pendulum 1 will gain over pendulum 2
Now comes the important thing - they will again be in phase when one of the pendulum GAINS ONE OSCILLATION over the other. In our case ,it will be pendulum 1 that gains over the other since its time period is lower and therefore it will be able to complete more oscillations than the other in a given time.

Let pendulum 1 gain one oscillation over 2 in time =t
Then no of oscillations by pendulum 1 in time t = n1= t/ T1 ......................(1)
no of oscillations by pendulum 2 in time t =n2 = t/ T2 ......................(2)

Therefore we require

n1-n2 =1

t/T1 - t/T2 = 1

From this equation we can find t.

Put t in eqn 1 and 2 and find the no of oscillations made by pendulum 1 and pendulum 2 in this given time.