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λ₁=50 pt and λ₂=60 pt

u(x,t) = 2A cos(1.0t-0.010x) sin(11.5t-0.115x) Let us first discuss the sinus part of the full disturbance. It contains 11.5×t which describes the fast oscillation with a period given by Tfast = 2π/11.5 = 0.55 s which is the average period of each of the two individual disturbances. You may verify that period by stopping the video (with the escape key). The length of the image equals 400 pt, whereas the speed of waves equals 100 pt/s. This implies that the length of the image represents 4 s. We count almost 8 cycles in the stopped image which corresponds thus to a bit more than 0.5 s per cycle. The sinus also contains 0.115×x. From this we may find the wavelength of the fast oscillation λfast = 2π/0.115 = 55 pt Indeed, almost 8 cycles in the 400 pt length of the stopped image imply a wavelength of a bit more than 50 pt.

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