Stroboscopic photography

The interval of seven moments in (1) graph is six, and the period is t = 0.1s× 6x2 =1.2s;

(2) The seven positions in the figure are symmetrical about the midpoint, and the amplitude is the maximum distance from the vibrator to the equilibrium position, so the distance between the two positions is twice the amplitude, which is 6cm, so the amplitude is 3 cm.

(3) By measuring the displacements S 1, S2, S3 ... the spring oscillators pass in adjacent equal time, we can know that the average speed of each displacement is. V = st, so we can know that in the process from A to B, the speed of the oscillator first increases and then decreases.

Answer: (1) The vibration period of this oscillator is1.2s; ; (2) The vibration amplitude of the vibrator is 3 cm; (3) In the process from A to B, the speed of the vibrator first increases and then decreases.