Particle speed maximizes at equilibrium in transverse waves

Steps:

• Water particles in a transverse wave undergo simple harmonic motion (SHM) vertically.
• Particle speed follows $v = \omega \sqrt{A^2 - y^2}$, where $y$ is vertical displacement from equilibrium.
• Speed is zero at maximum displacement ($y = \pm A$, crests/troughs) and maximum at $y = 0$.
• The graph's snapshot shows R at $y = 0$, P at a crest, and Q midway displaced.

Why D is correct:

• Particle R is at equilibrium ($y = 0$), maximizing speed per the SHM velocity formula.

Why the others are wrong:

• A: Speeds differ by position in SHM.
• B: Particle P at crest has $y = A$, so speed is zero.
• C: Particle Q has partial displacement, so speed is less than maximum.

Final answer: D