A particle of mass m is attached to a fixed point A by a model spring of natural

Question

A particle of mass m is attached to a fixed point A by a model spring of natural length I_0 and stiffness mg/l_0, and to another fixed point B vertically below A by another model spring, of natural length l_0 and stiffness 3mg/l_0, where g is the acceleration due to gravity. The distance between point A and B is 3l_0. The particle moves along the vertical line between A and B as shown. If m = 2.25 kg and I_0 = 0.517 m, calculate the equilibrium position in metres, relative to A. Give your answer to 3 decimal places, and take g = 9.81 ms^-2.

Explanation / Answer

equlibrium position, forces balance each other....

let it be at x distance from A.

so x-lo is stretching in upper spring.

so force = mg/lo * (x-lo)

mg force acts down.

for second spring , stretch = 2lo - x

also force of second spring = 3mg/lo * (2lo - x)

so mg/lo * (x-lo) = mg + 3mg/lo * (2lo - x)

x -lo = lo + 2lo - x

2x = 4 lo

x =2lo

x= 2*0.517 = 1.034 m

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