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Two identical spring-block systems (spring constant K, mass M each) are placed on a smooth horizontal table. Block A is displaced +a0 in the x-direction and Block B is displaced -a0/(2*sqrt(3)) in the x-direction. At t = 0, Block A is released from rest and Block B is given a velocity of (a0/2)*sqrt(K/M) in the +x direction. Then:

1. Time for both blocks to reach the same position for the first time is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K) sec.
2. Time for the distance between blocks to decrease to minimum is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K) sec.
3. Time for the relative velocity between blocks to become minimum is [pi/4 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K) sec.
4. The blocks will never be in the same phase.

Correct answer: Time for the distance between blocks to decrease to minimum is [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K) sec.

Solution

The relative displacement X_rel = x_A - x_B satisfies SHM with amplitude and phase found from initial conditions. Minimum distance occurs when relative velocity = 0, i.e., when X_rel reaches its minimum. Working through initial conditions gives t = [pi/2 + tan⁻¹(1/(3*sqrt(3)))] * sqrt(M/K).

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