Question

Why does the equilibrium position of the spring change when a mass is added to the spring? Will the mass oscillate around the new equilibrium position of the spring or the previous position without a mass attached to the spring? If the equilibrium position of the spring changes by 20 cm (assuming no initial mass) when a mass is added to the spring with constant 4.9 kg/s^2, what is the mass of the object attached to the spring?

Answer 1

Solution:

when a mass is added to the string its equilibrium position changes due to the wight of the mass added according to the formula,

=> mg = k*x

mass will oscillate about new equilibrium position.

given x = 0.2 m , k = 4.9

=> m * g = k * x

=> m = 4.9 * 0.2 / 9.8

=> m = 0.1 kg

thanks.