On a newly discovered planet, the radius of which is equal to the radius of the Earth, a ball suspended by a thread

On a newly discovered planet, the radius of which is equal to the radius of the Earth, a ball suspended by a thread performs small oscillations with a period of 4 seconds. When the thread was lengthened by 21 cm, the oscillation period became = 4.4 seconds. Find the lengths of the pendulums.

T1 = 4 s.

T2 = 4.4 s.

L2 – L1 = 21 cm = 0.21 m.

L1 -?

L2 -?

A ball suspended from a thread is a mathematical pendulum.

The period of oscillation of the mathematical pendulum T is the time of one complete oscillation.

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.

√L1 = T1 * √g / 2 * P.

L1 = T1 ^ 2 * g / 4 * P ^ 2.

L2 = T2 ^ 2 * g / 4 * P ^ 2.

L2 – L1 = T2 ^ 2 * g / 4 * P ^ 2 – T1 ^ 2 * g / 4 * P ^ 2 = (T2 ^ 2 – T1 ^ 2) * g / 4 * P2.

Let us express the acceleration of free fall on a new planet.

g = (L2 – L1) * 4 * P ^ 2 / (T2 ^ 2 – T1 ^ 2) = 2.47 m / s2.

g = 0.21 m * 4 * (3.14) ^ 2 / ((4.4 s) ^ 2 – (4 s) ^ 2).

L1 = (4 s) ^ 2 * 2.47 m / s2 / 4 * (3.14) ^ 2 = 1 m.

L2 = (4.4 s) ^ 2 * 2.47 m / s2 / 4 * (3.14) ^ 2 = 1.21 m.

Answer: L1 = 1 m, L2 = 1.21 m.