Two particles are in $SHM$ on same straight line with amplitude $A$ and $2A$ and with same angular frequency $\omega .$ It is observed that when first particle is at a distance $A/\sqrt{2}$ from origin and going toward mean position, other particle is at extreme position on other side of mean position. Find phase difference between the two particles
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Consider $2$ circles with radii $A \& 2 A$ and particles moving on them with constant angular velocity $\omega$
Choose $x$ axis to be horizontal on which the projection of the particles are to be taken.
When the projection of the first particle is at a distance $A / \sqrt{2}$ from origin, it is at an angle $45^{0}$ in the first quadrant.
whereas the second particle is at an angle $180^{0}$.
i.e the phase difference is $135^{0}$
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