Amplitude and wavelength from displacement-time graph

Steps:

• Amplitude is the maximum displacement from the equilibrium position, read directly from the graph's peak value as 3.0 units.
• Identify the period T as the time for one complete oscillation, measured between consecutive peaks on the graph.
• Calculate frequency f = 1/T, using the graph's time scale.
• Wavelength λ = c / f, where c = 3.0 × 10^8 m/s is the speed of light in vacuum, yielding λ = 6000 nm.

Why A is correct:

• Matches the graph's peak displacement of 3.0 units and calculated λ = 6000 nm using λ = c / f.

Why the others are wrong:

• B, C, D: Incorrect wavelength of 7500 nm, likely from misreading the period or frequency on the graph.

Final answer: A