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Two identical spring-block systems (spring constant K, mass M) are placed on a smooth horizontal table. At t = 0, block A is at displacement +a₀ from equilibrium and released from rest; block B is at displacement -a₀/(2*sqrt(3)) and given velocity a₀/2 * sqrt(K/M) in the +x direction. Which of the following statements is/are correct? (A) Time for both particles to first reach the same position is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K). (B) Time for the distance between them to first reach its minimum is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K). (C) Time for their relative velocity to first become minimum is [pi/4 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K). (D) The particles will never be in the same phase.

1. (A) Time for both particles to first reach the same position is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K).
2. (B) Time for the distance between them to first reach its minimum is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K).
3. (C) Time for their relative velocity to first become minimum is [pi/4 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K).
4. (D) The particles will never be in the same phase.

Correct answer: (B) Time for the distance between them to first reach its minimum is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K).

Solution

Block A oscillates as a₀*cos(omega*t). Block B has amplitude a₀/sqrt(3) with phase 4*pi/3, giving relative motion as a sinusoid whose first extremum (minimum separation) occurs at t = [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K). The particles share the same equilibrium and angular frequency so they can coincide (same position), but their phases differ such that minimum separation time matches option (B).

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