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Two particles oscillate in SHM along the same straight line, sharing the same mean position and the same angular frequency omega. Their amplitudes are A and 2A respectively. At a certain instant they are found at position x = A/2 from the mean position, moving in opposite directions. What is the phase difference between the two particles at that instant?

1. 5*pi/6 - sin⁻¹(1/4)
2. pi/6 - sin⁻¹(1/4)
3. 5*pi/6 - cos⁻¹(1/4)
4. pi/6 - cos⁻¹(1/4)

Correct answer: 5*pi/6 - sin⁻¹(1/4)

Solution

For opposite velocity directions, the cosines of the phase angles must have opposite signs; the consistent pairing gives phase difference delta = theta₁ - theta₂ = 5pi/6 - sin⁻¹(1/4).

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