A particle is making simple harmonic motion along the X-axis. If at a distances x₁ and x₂ from the mean position the velocities of the particle are v₁ and v₂ respectively. The time period of its oscillation is given as :

Solution

Physics sir · 3 steps

Write v² = ω²(A² - x²) at both points and subtract, so the unknown amplitude drops out.

1. Velocity in SHM. At a distance x from the mean position, the speed is v = ω√(A² - x²), so v² = ω²(A² - x²). We do not know the amplitude A, so we must get rid of it.
2. Two points, subtract. Write v₁² = ω²(A² - x₁²) and v₂² = ω²(A² - x₂²). Subtracting the second from the first cancels ω² A² and leaves v₁² - v₂² = ω²(x₂² - x₁²).

   ω = √((v₁² - v₂²)/(x₂² - x₁²)) ⇒ T = 2π/ω = 2π√((x₂² - x₁²)/(v₁² - v₂²))

3. So. The time period matches option (d). The answer makes sense: if x₂ > x₁, the particle is farther out and slower, so both differences are positive.

The idea to keep. When the amplitude is unknown, use v² = ω²(A² - x²) at two positions and subtract. A cancels, and both differences come with matching signs.

Answer from the NTA final answer key. Solution by PrepWiser.