Question

Two points are moving with simple harmonic motions of same period and amplitude along the same line. They are found to cross each other in opposite directions, always at the point for which the displacement $x=\frac{A}{2}$, where A is the amplitude. Find the “phase difference” between the two motions.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The two points cross each other at M (here OM = x = $\frac{A}{2}).$ Time taken for the first particle to move from “O” to “M” is T/12.

The time taken by the other particle to move from “O” to “M” vial “X” will be $\frac{T}{2}-\frac{T}{12}=\frac{5T}{12}.$
Therefore, the time difference is $\frac{5T}{12}-\frac{T}{12}=\frac{T}{3}.$
The corresponding phase difference is $\frac{2\pi }{3}$. (T corresponds to a phase difference of $2\pi$)