Wave speed from amplitude, period, and wavelength

Steps:

• Identify amplitude A from graph as half the peak-to-peak height (e.g., 2 cm).
• Determine period T from one full cycle time on graph (e.g., 0.5 s).
• Calculate frequency f = 1/T (e.g., 2 Hz).
• Compute wave speed v = f λ, but since numerical values of A and λ are equal (e.g., both 2 cm), and T provides f, v = λ / T = 2 cm / 0.5 s = 4 cm/s.

Why C is correct:

• Wave speed v = f λ = (1/T) λ; with A = λ numerically (both 2 cm) and T = 0.5 s, v = 4.0 cm/s matches the formula.

Why the others are wrong:

• A: Assumes T = 2 s, doubling period halves speed incorrectly.
• B: Assumes T = 1 s, yielding v = 2 cm/s but ignores graph's shorter cycle.
• D: Assumes T = 0.25 s, halving period doubles speed beyond graph data.

Final answer: C