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.

**the figure is a spring hanging vertically with two masses attached. Between the masses is point c.**

Part A: What is the equilibrium distance of point A when both masses are attached as shown?

Part B: If only the upper mass is attached, what is the equilibrium distance of point A?

Part C: 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.

Part D: Find the amplitude of the motion of the remaining mass.

Part E: What is the maximum velocity of the remaining mass during its oscillation?

Explanation / Answer

A - for equilibrium distance of point A when both masses are attached , kx = mg

x= mg/k

or, x = 2*9.8/100

= 19.6 cm

equilibrium distance = 10 +x = 29.6 cm

B- If only the upper mass is attached ,

, kx = mg

x= mg/k

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