Two mathematical pendulums with lengths l1 and l2 simultaneously begin oscillations in the same phases

Two mathematical pendulums with lengths l1 and l2 simultaneously begin oscillations in the same phases with periods T1 = 6s and T2 = 5s. After what minimum time will the phases of their oscillations be the same again? What will be the oscillation frequency f of a pendulum with a length L = l1 + l2?

The time t after which the oscillation phases of the pendulums will be the same again is equal to the smallest common multiple of their periods. In this case, the least common multiple is equal to the product of periods:

t = 5 * 6 = 30 s.

Dependence of the periods of oscillations on the length of the pendulums:

T1 = 2п√ (l1 / g);

T2 = 2п√ (l2 / g).

We find the squares of the periods and the values ​​of the lengths of the pendulums:

T12 = 4n ^ 2 (l1 / g);
T12 = (4n ^ 2 / g) l1;
l1 = (T1 ^ 2 * g) / (4n ^ 2).

Similarly:

l2 = (T2 ^ 2 * g) / (4n ^ 2);

L = l1 + l2 = (T1 ^ 2 * g) / (4n ^ 2) + (T2 ^ 2 * g) / (4n ^ 2) = (g / (4n ^ 2)) * (T1 ^ 2 + T2 ^ 2);

The period T of oscillations of a pendulum with a length L:

T = 2п √ (((g / (4п ^ 2)) * (T1 ^ 2 + T2 ^ 2)) / g) =
= √ (T1 ^ 2 + T2 ^ 2) = √ (36 s2 + 25 s2) = √61 s ≈ 7.811 s.

Answer. The phases will coincide again after 30 seconds. The oscillation period of the composite pendulum is T = √61 s ≈ 7.811 s.