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In this this question, the displacement-time graphs are given, which are different from displacement-distance graphs.
In the displacement-time graphs of A and B, they show the displacement of that particular point at different timing. E,g, at t = 0s, the A is at the rest position (0 displacement) and at time 0.2 s it is at the maximum displacement. This means A is going up from t = 0 s to 0.2 s.
Solutions:
(a) Amplitude: 1.5 cm
(b)(i) Frequency is the number of complete waves produced in 1 second.
(ii) period T = 0.8s, f = 1/T = 1/0.8 = 1.25 Hz
(c)(i) Closest possible positions of A and B, (refer to the video), is when the time taken for the wave to move from A to B is T/4 = 0.8/4 = 0.2 s.
speed = distance/time = 38/0.2 = 190 cm/s approx. 200 cm/s
(ii) There are various possibilities in which B can be 38 cm to the right of A. Besides T/4, it can be 1.25T or 2.25 T etc. Hence the speed can be other values.
Refer to the video explanation below
