In a time interval of 32 s, one of the mathematical pendulums performed the same number

In a time interval of 32 s, one of the mathematical pendulums performed the same number of complete oscillations as the other in a time interval of 64 s. find the lengths of the pendulums if one of them is shorter than the other by 33 cm

t1 = 32 s.

t2 = 64 s.

N1 = N2.

L2 – L1 = 33 cm = 0.33 m.

g = 9.8 m / s2.

L1 -?

L2 -?

The period of oscillation of the 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 = t1 / N1.

T2 = t2 / 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’s thread, g is the acceleration of gravity.

T1 = 2 * P * √L1 / √g.

t1 / N1 = 2 * P * √L1 / √g.

t1 ^ 2 / N1 ^ 2 = 4 * P ^ 2 * L1 / g.

L1 = g * t1 ^ 2/4 * P ^ 2 * N1 ^ 2.

L2 = g * t2 ^ 2/4 * P ^ 2 * N2 ^ 2.

L1 = g * t1 ^ 2/4 * P ^ 2 * N1 ^ 2.

L2 = g * t2 ^ 2/4 * P ^ 2 * N2 ^ 2.

L2 – L1 = g * t2 ^ 2/4 * P ^ 2 * N2 ^ 2 – g * t1 ^ 2/4 * P ^ 2 * N1 ^ 2 = g * (t2 ^ 2 – t1 ^ 2) / 4 * N ^ 2 * N1 ^ 2.

N1 ^ 2 = g * (t2 ^ 2 – t1 ^ 2) / (L2 – L1) * 4 * P ^ 2.

N12 = 9.8 m / s2 * ((64 s) ^ 2 – (32 s) ^ 2) / 0.33 m * 4 * (3.14) ^ 2 = 2304.

N1 = 48.

L1 = 9.8 m / s2 * (32 s) ^ 2/4 * (3.14) ^ 2 * (48) ^ 2 = 0.11 m.

L2 = 9.8 m / s2 * (64 s) ^ 2/4 * (3.14) ^ 2 * (48) ^ 2 = 0.44 m.

Answer: L1 = 0.11 m, L2 = 0.44 m.