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As shown in the figure, a and B balls are attached to the lower end of the vertical suspension light spring, and their mass ma = 0.1kg and MB = 0.5kg. When the spring is at rest, its elongation is 15cm. If the thin line between a and B is cut, what is the amplitude and maximum acceleration of a ball in simple harmonic motion?
According to the force balance condition of two balls at rest, the stiffness coefficient of spring is obtained as follows: KX = (MA + MB) g; k = (MA + MB) GX = 40n / m, after cutting the thin line between a and B, the elongation of spring when a ball is still suspended is XA = magk = 0.025m, and this position at the lower end of spring is the equilibrium position in the vibration of a ball When the spring is more extended than the static suspension a ball, the length is the amplitude, that is, a = x-xa = 15cm-2.5cm = 12.5cm, the maximum acceleration of a ball in vibration is am = Kam = 50M / S2; a: the maximum amplitude is 12.5cm; the maximum acceleration is 50M / S2
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