Two particles executes shm of same amplitude and frequency along the same straight line They pass one another when going in opposite directions each time their displacement is half of their amplitude The phase difference between them is

Dear Pallavi,

We can solve this problem graphically. Now consider the graph as shown in the figure this would be the same for the other case too with a phse shift.

Now for one of the displacement would be half of its amplitude twice in a possitive cycle i.e. from 0 to ∏. The two pionts are  when pase term ωt is ∏/6 and 5∏/6. So for the two objects under given conditions .i.e one is going in the direction of its amplitude increase(0 to ∏/2) and the other is in the direction of its amplitude decrease i.e. from ∏/2 to ∏.

So the minimum phase difference is 5∏/6 - ∏/6 = 2∏/3

this phase difference may be even multiples of ∏.

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