Velocity in simple harmonic motion (SHM) For a particle executing SHM with amplitude r and instantaneous displacement y from the mean position, which expression gives its speed v at that instant?

Introduction / Context:Simple harmonic motion (SHM) models oscillations of springs, pendulums (small angles), and many vibrating systems. The velocity–displacement relation is essential for predicting speeds at different positions and for energy-based design.

Given Data / Assumptions:

• Amplitude r (maximum magnitude of displacement).
• Instantaneous displacement y from the mean (equilibrium) position.
• Angular frequency ω (rad/s) characterizes the system.

Concept / Approach:For SHM: y = r * sin(ω t + φ). Differentiate to get velocity: v = dy/dt = r * ω * cos(ω t + φ). Using the identity sin^2θ + cos^2θ = 1 and eliminating time gives v^2 = ω^2 (r^2 − y^2), hence v = ω * sqrt(r^2 − y^2). This highlights maximum speed at y = 0 and zero speed at y = ±r.

Step-by-Step Solution:

Verification / Alternative check:Energy method: total energy E = (1/2) k r^2 is constant. At displacement y, elastic potential is (1/2) k y^2, so kinetic K = E − U = (1/2) k (r^2 − y^2). Using k = m ω^2 and K = (1/2) m v^2 gives v = ω * sqrt(r^2 − y^2).

Why Other Options Are Wrong:

• ω^2(r − y), ω(r + y), ω y: Do not satisfy boundary conditions (e.g., v = 0 at y = ±r).
• sqrt(ω^2 r^2 + y^2): Speed would increase with |y|, contradicting SHM behaviour.

Common Pitfalls:Forgetting that maximum speed occurs at the mean position and misusing signs when eliminating time.

Final Answer:v = ω * sqrt(r^2 - y^2)