Question

Two particles are in SHM on a 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.

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${135}^{\circ }$
• D. ${180}^{\circ }$

Answer 1

Answer: (C)

${135}^{\circ }$

Consider 2 circles with radii $A$ & $2A$ 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 ${135}^0$