A particle executes simple harmonic motion between x=-A and x=+A. If time taken by particle to go from x=0 to A/2 is 2 s; then time taken by particle in going from x=A/2 to A is

Solution

Physics sir · 4 steps

Do not split the distance, split the phase: the angle swept tells you the time.

1. Start from mean. Measure time from x = 0, so the position is x = A sin ω t. The phase ω t grows at a steady rate, so equal phase changes take equal times.

   A/2 = A sin ω t ⇒ ω t = π/6

2. Find the period. A phase of π/6 is one twelfth of a full cycle 2π, so going from 0 to A/2 takes T/12 = 2 s. That gives T = 24 s.
3. Second part. At x = A the phase is π/2. From A/2 to A the phase changes by π/2 - π/6 = π/3, which is one sixth of a cycle: T/6 = 24/6 = 4 s.
4. So. The particle slows down near the extreme, so the second half of the distance takes twice as long: 4 s, option (a).

The idea to keep. In SHM, convert positions into phase using x = A sin ω t. Time is proportional to phase: 0 to A/2 is T/12, and A/2 to A is T/6.

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