A particle oscillates with simple harmonic motion. Its displacement from equilibrium is given by x(t) = 0.12 sin(4π t + π/6), where x is in meters and t is in seconds. What is the maximum speed of the particle?

Correct answer

A. 0.48π m/s

Principle or equation

For simple harmonic motion x(t) = A sin(ωt + φ), the velocity is v(t) = Aω cos(ωt + φ). The maximum speed is v_max = Aω.

Why this answer is correct

Here amplitude A = 0.12 m and angular frequency ω = 4π rad/s. Thus v_max = Aω = 0.12 × 4π = 0.48π m/s.

Example

If A = 0.10 m and ω = 2π rad/s, then v_max = 0.10 × 2π = 0.20π m/s.