A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.

Solution

Physics sir · 3 steps

Start the clock at the extreme and use cosine: half amplitude comes at a phase of exactly π/3.

1. Choose the equation. The particle starts at maximum displacement, so take x = A cos ω t. This gives x = A at t = 0.
2. Angular frequency. ω = 2π/T = 2π/6 = π/3 rad s⁻¹.

   A/2 = A cos(π/3t) ⇒ cos(π/3t) = 1/2 ⇒ π/3t = π/3

3. So. t = 1 s, which is T/6. The amplitude of 8 cm is not needed, because only the ratio x/A = 1/2 matters. The answer is 1.

The common slip: 0.5. Using x = A sin ω t gives 0.5 s, but that is the time from the mean position to A/2, not from the extreme.

The idea to keep. Pick the equation that matches the starting point: cosine from an extreme, sine from the mean. Extreme to A/2 takes T/6, while mean to A/2 takes T/12.

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