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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) The cosine part is shown in the above video. The cosine part of the full disturbance contains 1.0×t which describes the slow oscillation with a period given by Tfast = 2π/1.0 = 6.28 s which is the difference in period of each of the two individual disturbances. Since the length of the image represents 4 s it contains a little bit more than half a cycle of the slow oscillation. The cosine part also contains 0.010×x. From this we may find the wavelength of the fast oscillation λfast = 2π/0.010 = 628 pt Indeed, only a bit more than half a cycle fits in the 400 pt length of the video.

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