# What is the relation of the masses of two spring pendulums, oscillating on the same springs

**What is the relation of the masses of two spring pendulums, oscillating on the same springs, if during the same time the first one made 10 oscillations, and the second 40 oscillations?**

N1 = 10.

N2 = 40.

t1 = t2.

k1 = k2.

m1 / m2 -?

The period of oscillation of the pendulum T is the time of one complete oscillation: T = t / N, where t is the oscillation time, N is the number of oscillations.

T1 = t1 / N1.

T2 = t2 / N2.

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

T1 = 2 * P * √m1 / √k1.

T2 = 2 * P * √m2 / √k2.

2 * P * √m1 / √k1 = t1 / N1.

4 * P ^ 2 * m1 / k1 = t1 ^ 2 / N1 ^ 2.

m1 = k1 * t1 ^ 2/4 * P ^ 2 * N1 ^ 2.

m2 = k2 * t2 ^ 2/4 * P ^ 2 * N2 ^ 2.

m1 / m2 = 4 * P ^ 2 * N2 ^ 2 * k1 * t1 ^ 2/4 * P ^ 2 * N1 ^ 2 * k2 * t2 ^ 2 = N2 ^ 2 / N1 ^ 2.

m1 / m2 = (40) ^ 2 / (10) ^ 2 = 16.

Answer: the mass of the first cargo m1 is 16 times greater than the second m2: m1 / m2 = 16.