Under the action of a load weighing 100 g, the spring was deformed by 1 cm. Determine the weight of the load that

Under the action of a load weighing 100 g, the spring was deformed by 1 cm. Determine the weight of the load that must be fixed on the spring so that the period of its oscillations is 1 s?

m1 = 100 g = 0.1 kg.

g = 10 m / s2.

x1 = 1 cm = 0.01 m.

T2 = 1 s.

m2 -?

The period of free oscillations of a spring pendulum T is determined by the formula: T = 2 * п * √m / √k, where п is the number pi, m is the mass of the load, k is the stiffness of the spring.

T2 = 2 * п * √m2 / √k.

m2 = T2 ^ 2 * k / 4 * P ^ 2.

Let us write down the equilibrium condition when suspending a load of mass m1: m1 * g = k * x1.

k = m1 * g / x1.

m2 = T2 ^ 2 * m1 * g / 4 * x1 * п ^ 2.

m2 = (1 s) ^ 2 * 0.1 kg * 10 m / s2 / 4 * 0.01 m * (3.14) ^ 2 = 2.54 kg.

Answer: on the spring it is necessary to secure a weight of m2 = 2.54 kg.