A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s. The maximum velocity of the particle is _________ cm / s.

Solution

Physics sir · 3 steps

With T = 3.14 s, the angular frequency comes out to exactly 2, so the speed follows at once.

1. Angular frequency. For SHM, ω = 2π/T. With T = 3.14 s and π ≈ 3.14, this gives ω = (2 × 3.14)/3.14 = 2 rad s⁻¹.
2. Maximum speed. The particle is fastest at the mean position, where all the energy is kinetic. There the speed is the amplitude times the angular frequency.

   vₘₐₓ = Aω = 0.06 m × 2 s⁻¹ = 0.12 m s⁻¹

3. So. Converting to cm, 0.12 m s⁻¹ = 12 cm s⁻¹, so the answer is 12.

The common slip: 0.12. Leaving the answer in metres per second gives 0.12, not 12. Multiply by 100 to convert to cm/s.

The idea to keep. In SHM the top speed is always Aω, reached at the mean position. Find ω = 2π/T first, then watch the units the answer asks for.

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