Question

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point where the spring is not stretched. Then the mass is released and begins to oscillate. The lowest point in the oscillation is 18 cm below the point where the mass was released. What is the oscillation frequency?

Answer 1

When the mass falls a distance ∆x under gravity is loses potential energy equal to m*g*∆x. That energy is gained by the spring and is equal to 0.5*k*∆x². At the lowest point of oscillation there is no kinetic energy, so the spring stored energy and the gravitational potential energy are equal

m*g*∆x = 0.5*k*∆x²

m*g = 0.5*k*∆x

solve for k

k = 2*m*g/∆x

The resonant frequency of a spring-mass system is (1/2π)*√[k/m]; substituting for k:

f0 = (1/2π)*√[2*m*g/(m*∆x)] = (1/2π)*√[2*g/∆x]

∆x = 0.18 m, g = 9.8 m
f_{​​​​0} = (1/2π) * √(2*9.8/0.18)
f0 = 1.66Hz

I hope it was helpful.. please rate me positive