simple harmonic oscillator has a period of 0.02s and amplitude of 0.25m. Calculate the speed in $ms^{-1}$ at the centre of the oscillation

Analyze the question: We need the speed at the centre of a simple harmonic oscillation. At the centre (the equilibrium position), the speed is at its maximum.

Concept:

• Angular frequency: ω = 2π/T
• Maximum speed: v_max = ωA (from v = ω√(A² − x²), with x = 0)

Solution:

1. ω = 2π / T = 2π / 0.02 = 100π rad/s

2. v_max = ωA = 100π × 0.25 = 25π m/s

Check against the options: 25π matches option B (25.00π).

Answer: The speed at the centre is 25π m/s (about 78.5 m/s).

Tip: At the centre (x = 0), speed is maximum and acceleration is zero. At the extreme ends (x = ±A), speed is zero and acceleration is maximum. Remember v_max = ωA = 2πA/T, and you can solve this kind of question in two lines.