Two pendulums, the lengths of which differ by 22 cm, perform in the same place on the Earth

Two pendulums, the lengths of which differ by 22 cm, perform in the same place on the Earth during some time one – 30 vibrations, the other – 36 vibrations. Find the lengths of the pendulums.

The number of oscillations of pendulums with periods T1 and T2 during time t:

n1 = t / T1;

n2 = t / T2;

T1 = 2pi√ (L / g) (L is the length of the thread, g is the acceleration of gravity);

T2 = 2pi√ ((L + 0.22) / g);

n1 / n2 = (t / T1) / (t / T2) = T2 / T1 = (2pi√ ((L + 0.22) / g)) / 2pi√ (L / g);

36/30 = √L / √ (L + 0.22);

6/5 = √L / √ (L + 0.22);

36/25 = L / (L + 0.22);

36 (L + 0.22) = 25L;

36L + 5.5 = 25L;

11L = 5.5;

L = 5 m.

L + 0.22 m = 5 m + 0.22 m = 5.22 m.

Pendulum lengths: 5 m and 5.22 m.