Simple harmonic motion (SHM) application: For a body executing SHM with amplitude 150 mm and frequency 2 vibrations/s, what is the maximum (peak) speed attained during the motion?

Introduction / Context:

This problem checks recall and correct use of simple harmonic motion (SHM) kinematics. In SHM, the displacement, velocity, and acceleration are sinusoidal functions, and the peak (maximum) speed depends on the angular frequency and the amplitude. Engineers use this relation for designing vibration isolators, cams, and reciprocating systems.

Given Data / Assumptions:

• Amplitude A = 150 mm = 0.150 m.
• Frequency f = 2 vibrations/s.
• Angular frequency ω = 2 * π * f.
• Ideal SHM with no damping.

Concept / Approach:

The SHM velocity varies as v(t) = ω * A * cos(ωt + φ). The maximum magnitude occurs when |cos(…)| = 1, hence v_max = ω * A. Convert frequency to angular frequency and multiply by amplitude.

Step-by-Step Solution:

Verification / Alternative check:

Using energy: at mid-position, all energy is kinetic: (1/2) m v_max^2 = (1/2) k A^2 with k = m * ω^2, giving v_max = ω * A again. This independently confirms the result.

Why Other Options Are Wrong:

• 0.18845 m/s and 0.01885 m/s are off by factors of 10 and 100 due to unit/frequency mistakes.
• 18.85 m/s and 188.5 m/s are 10× and 100× too large, likely from using A in mm without converting to metres or squaring π incorrectly.

Common Pitfalls:

• Forgetting to convert amplitude from mm to m.
• Using frequency f instead of angular frequency ω in v_max = ω * A.

Final Answer:

1.885 m/s