A particle executes 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:
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A particle executes 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:

G1 Physics · Accepted Answer

Step 1: The SHM speed–displacement relation is v² = ω²(A² − x²). Step 2: At x₁ and x₂: v₁² = ω²(A² − x₁²), v₂² = ω²(A² − x₂²). Step 3: Subtract: v₁² − v₂² = ω²(x₂² − x₁²). Step 4: Hence ω² = (v₁² − v₂²)/(x₂² − x₁²), and T = 2π√[(x₂² − x₁²)/(v₁² − v₂²)]. Final answer: (B) 2π√[(x₂² − x₁²)/(v₁² − v₂²)].