Two weights are alternately suspended from the spring. The period of oscillation of 1 load is T = 0.6 s

Two weights are alternately suspended from the spring. The period of oscillation of 1 load is T = 0.6 s, the second is T = 0.8 s. What will be the period of oscillation if 2 weights are simultaneously suspended from this spring?

T1 = 0.6 s.

T2 = 0.8 s.

m = m1 + m2.

T – ?

The period of oscillation of the pendulum T is the time of one complete oscillation. For a spring pendulum, the period of free natural oscillations 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 / √k.

√m1 = T1 * √k / 2 * P.

m1 = T1 ^ 2 * k / 4 * P ^ 2.

T2 = 2 * P * √m2 / √k.

√m2 = T2 * √k / 2 * P.

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

T = 2 * P * √ (m1 + m2) / √k = 2 * P * √ (T1 ^ 2 * k / 4 * P ^ 2 + T2 ^ 2 * k / 4 * P ^ 2) / √k = 2 * P * √ (T1 ^ 2/4 * P ^ 2 + T2 ^ 2/4 * P2).

T = 2 * 3.14 * √ ((0.6 s) ^ 2/4 * (3.14) ^ 2 + (0.8 s) ^ 2/4 * (3.14) ^ 2) = 1 with.

Answer: the period of free natural oscillations will become T = 1 s.