Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance X0(X0>A). If the maximum separation between them is (X0+A), the phase difference between their motion is:
(A)3π✓ Correct
(B)4π
(C)6π
(D)2π
Answer: (A)
Solution
Take x₁ = A sin ωt and x₂ = X₀ + A sin(ωt + φ). Their separation is X₀ + A[sin(ωt+φ) − sin ωt] = X₀ + 2A sin(φ/2) cos(ωt + φ/2), whose largest value is X₀ + 2A sin(φ/2). Setting this equal to the given maximum X₀ + A gives 2A sin(φ/2) = A, so sin(φ/2) = 1/2 and φ/2 = π/6. Therefore φ = π/3.
JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option
