Two particles execute SHM with amplitudes A and 2A and angular frequencies ω and 2ω. At t = 2π/(3ω) they are in the same phase. If the initial phase difference is 2π/n, find n.
Physics · Simple Harmonic Motion
JEE Mains

Question

Two particles execute SHM with amplitudes A and 2A and angular frequencies ω and 2ω. At t = 2π/(3ω) they are in the same phase. If the initial phase difference is 2π/n, find n.

G1 Physics · Accepted Answer

The relative phase evolves at rate (2ω−ω)=ω. If the initial phase difference is 2π/n and they become in phase at t=2π/(3ω), then the relative phase changes by 2π/3. The smallest positive initial difference that is cancelled is 2π/3, so 2π/n=2π/3 and n=3.