The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :

Solution

Physics sir · 3 steps

Starting from the mean position means a sine; half amplitude means the phase is π/6, and that fixes T.

1. Pick the equation. The oscillator starts at the mean position, where x = 0 at t = 0. So the displacement is x = A sin ω t with ω = 2π/T.
2. Use the condition. At t = 3 s, x = A/2, so sin(ω × 3) = 1/2. The first time this happens is when the phase is π/6.

   2π/T × 3 = π/6 ⇒ T = 36 s

3. So. The time period is 36 s, which is option (d). Reaching half the amplitude takes only T/12, not T/8, because the particle moves fastest near the mean position.

The idea to keep. In SHM, displacement is not proportional to time. From the mean position, use x = A sin ω t. Reaching A/2 takes T/12 and reaching A takes T/4.

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