How many times is the potential energy accumulated by a spring when compressed from the equilibrium

How many times is the potential energy accumulated by a spring when compressed from the equilibrium position by 2 cm less than when the same spring is compressed by 4 cm?

k1 = k2 = k – the coefficient of stiffness of the springs is the same;

dx1 = 2 centimeters = 0.02 meters – the amount of compression of the spring in the first case;

dx2 = 4 centimeters = 0.04 meters – the amount of compression of the spring in the second case.

It is required to determine E1 / E2 – how many times the potential energy of the spring in the first case is less than in the second.

The potential energy of the spring in the first case will be equal to:

E1 = k1 * dx1 ^ 2/2 = k * dx1 ^ 2/2.

The potential energy of the spring in the second case will be equal to:

E2 = k2 * dx2 ^ 2/2 = k * dx2 ^ 2/2.

Then:

E1 / E2 = (k * dx1 ^ 2/2) / (k * dx2 ^ 2/2) = dx1 ^ 2 / dx2 ^ 2 = 0.02 ^ 2 / 0.04 ^ 2 = 0.0004 / 0, 0016 = 0.25, that is, E1 is 4 times less than E2.

Answer: 4 times less.