A particle oscillates with simple harmonic motion. Its displacement from equilibrium is given by x(t) = 0.080 sin(5π t + π/3), where x is in meters and t is in seconds. What is the maximum speed of the particle?
Show the answer and explanation
Correct answer
B. 1.26 m/s
Principle or equation
In simple harmonic motion, x(t) = A sin(ωt + φ), the velocity is v(t) = Aω cos(ωt + φ), so the maximum speed is v_max = Aω.
Why this answer is correct
From the equation, amplitude A = 0.080 m and angular frequency ω = 5π rad/s. Thus v_max = Aω = 0.080 × 5π = 0.080 × 15.708 = 1.257 m/s, approximately 1.26 m/s.
Example
If A = 0.10 m and ω = 2π rad/s, then v_max = 0.10 × 2π = 0.628 m/s.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.