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A particle is executing SHM along a straight line. Its velocities at distances x₁ and x₂ from the mean position are V₁ and V₂ respectively. Its time period is

1. 2π √(V₁² - V₂²) / (x₁² - x₂²)
2. 2π √(V₁² + V₂²) / (x₁² + x₂²)
3. 2π √(V₁² - V₂²) / (x₁² + x₂²)
4. 2π √(V₁² + V₂²) / (x₁² - x₂²)

Correct answer: 2π √(V₁² + V₂²) / (x₁² - x₂²)

Solution

In SHM, the velocity is given by v = √(ω²(A² - x²)), where ω is the angular frequency and A is the amplitude. Using this relation for two positions x₁ and x₂, and solving for the time period T = 2π/ω, we find the correct expression matches option D.

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