A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is 1/as. The value of 'a' to the nearest integer is _________.

Solution

Physics sir · 3 steps

Start the clock at the mean position, so use sine, and ask what phase gives half the amplitude.

1. Angular frequency. With T = 2 s, ω = 2π/T = π rad s⁻¹.
2. Choose the equation. The particle starts at the mean position, where x = 0 at t = 0. So the displacement is x = A sin ω t.

   A/2 = A sin(π t) ⇒ sin(π t) = 1/2 ⇒ π t = π/6

3. So. t = 1/6 s, which matches 1/a s, so a = 6.

The common slip: 3. Using x = A cos ω t measures time from the extreme position, giving π t = π/3, so t = 1/3 s and a = 3.

The idea to keep. Pick sine when motion starts at the mean position and cosine when it starts at an extreme. From the mean, reaching A/2 takes T/12. From an extreme, reaching A/2 takes T/6.

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