A load suspended on a spring with a stiffness of 250N / m makes 50 vibrations in 3.14 s

A load suspended on a spring with a stiffness of 250N / m makes 50 vibrations in 3.14 s to determine the weight of the load.

k = 250 N / m.

n = 50.

t = 3.14 s.

m -?

The frequency of oscillations of a spring pendulum v is the number of oscillations of the pendulum per unit of time, that is, in a time of 1 s.

The oscillation frequency v is determined by the formula: v = n / t, where n is the number of oscillations, t is the time during which the spring pendulum oscillates.

The frequency of a spring pendulum v is determined by a different formula: v = √k / 2 * P * √m, where k is the stiffness of the spring, P is the number pi, and m is the mass of the load.

n / t = √k / 2 * P * √m.

√m = t * √k / 2 * P * n.

m = t ^ 2 * k / 4 * P ^ 2 * n ^ 2.

m = (3.14 s) ^ 2 * 250 N / m / 4 * (3.14) ^ 2 * (50) 2 = 0.025 kg.

Answer: the mass of the cargo is m = 0.025 kg.