During the same period of time, one mathematical pendulum made N1 = 30 oscillations

During the same period of time, one mathematical pendulum made N1 = 30 oscillations, and the other -N2 = 36 oscillations. Find their lengths if one of them is shorter than the other by l = 22cm

N1 = 30.

N2 = 36.

t1 = t2.

l1 – l2 = 22 cm = 0.22 m.

l1 -?

l2 -?

T1 = t1 / N1.

T2 = t1 / N2.

T1 = 2 * P * √l1 / √g.

T2 = 2 * P * √l2 / √g.

t1 / N1 = 2 * P * √l1 / √g.

t2 / N2 = 2 * P * √l2 / √g.

l1 = g * t1 ^ 2 / N1 ^ 2 * 4 * P ^ 2.

l2 = g * t2 ^ 2 / N2 ^ 2 * 4 * P ^ 2.

l1 – l2 = g * t1 ^ 2 / N1 ^ 2 * 4 * P ^ 2 – g * t2 ^ 2 / N2 ^ 2 * 4 * P ^ 2 = g * t1 ^ 2 * (N2 ^ 2 – N1 ^ 2 ) / (N1 ^ 2 * N2 ^ 2 * 4 * P ^ 2).

t1 = √ ((l1 – l2) * (N1 ^ 2 * N2 ^ 2) * 4 * P ^ 2 / g * (N2 ^ 2 – N1 ^ 2)).

t1 = √ (0.22 m * ((30) ^ 2 * (36) ^ 2) * 4 * 3.14 ^ 2 / 9.8 m / s ^ 2 * ((36) ^ 2 – (30) ^ 2) = 51 s.

l1 = 9.8 m / s ^ 2 * (51 s) ^ 2 / (30) ^ 2 * 4 * (3.14) ^ 2 = 0.72 cm.

l2 = 0.5 m.

Answer: l1 = 0.72 cm, l2 = 0.5 m.