A spring with force constant k = 100 N/m has two masses hanging from it, as draw

Question

A spring with force constant k = 100 N/m has two masses hanging from it, as drawn (M = 1 kg) (Figure 1) . Without the masses, the free end of the spring (point A) is at an equilibrium distance of 10 cm below its support point.

1.Suppose that both masses are present and the system is initially motionless. If the string connecting the two masses is suddenly cut (at point C in the diagram) and the lower mass falls to the floor, find the period of the resulting oscillation of the remaining mass.

2.What is the maximum velocity of the remaining mass during its oscillation?

Explanation / Answer

T = 2 * pi / w = 2 * pi / 10 = .628 sec
2 M g = k x1    where x1 is initial elongation of spring
x1 = 2 * 9.8 / 100 = .196 m
E = 1/2 k x1^2       total energy stored in spring
E = 1/2 * 100 * .196^2 = 1.92 J
1/2 m v^2 = 1.92 J       maximum KE
v = (2 * 1.92)^1/2 = 1.96 m/s    maximum velocity

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