The pendulum made 120 oscillations in 2 minutes. When the length of the pendulum was increased by 74.7

The pendulum made 120 oscillations in 2 minutes. When the length of the pendulum was increased by 74.7, it made 60 oscillations in the same time. Find the starting and ending lengths of the pendulum and the acceleration due to gravity at that location.

t = 2 min = 120 s.
N1 = 120.
N2 = 60.
L2 = L1 + 0.747.
L1 -?
L2 -?
g -?
The period of oscillation of any pendulum T is the time of one complete oscillation. The oscillation period T is determined by the formula: T = t / N, where t is the time during which the pendulum makes N complete oscillations.
T1 = t / N1.
T2 = t / N2.
The period of the mathematical pendulum T is determined by the formula: T = 2 * P * √L / √g, where P are the numbers pi, L is the length of the pendulum thread, g is the acceleration of gravity.
T1 = 2 * P * √L1 / √g.
t / N1 = 2 * P * √L1 / √g.
t2 / N1 ^ 2 = 4 * P ^ 2 * L1 / g.
L1 = g * t ^ 2/4 * P ^ 2 * N1 ^ 2.
L2 = g * t ^ 2/4 * P ^ 2 * N2 ^ 2.
L1 = g * t ^ 2/4 * P ^ 2 * N1 ^ 2.
L2 = g * t ^ 2/4 * P ^ 2 * N2 ^ 2.
L2 – L1 = 0.747.
L2 – L1 = g * t2 / 4 * P ^ 2 * N2 ^ 2 – g * t ^ 2/4 * P ^ 2 * N1 ^ 2 = g * t ^ 2 * (N1 ^ 2 – N2 ^ 2) / 4 * P ^ 2 * N1 ^ 2 * N2 ^ 2.
g = (L2 – L1) * 4 * P ^ 2 * N1 ^ 2 * N2 ^ 2 / t ^ 2 * (N1 ^ 2 – N2 ^ 2).
g = 0.747 * 4 * (3.14) ^ 2 * (120) ^ 2 * (60) ^ 2 / (120 s) ^ 2 * ((120) ^ 2 – (60) ^ 2) = 9.821 m / c2.
L2 – L1 = 9.821 m / s2 * (120 s) ^ 2 * ((120) ^ 2 – (60) ^ 2) / 4 * (3.14) ^ 2 * (120) ^ 2 * (60) ^ 2 = 0.747 m.
L1 = 9.821 m / s2 * (120 s) ^ 2/4 * (3.14) ^ 2 * (120) ^ 2 = 0.249 m.
L2 = 0.249 m + 0.747 m = 0.996 m.
Answer: L1 = 0.249 m, L2 = 0.996 m, g = 9.821 m / s2.