Determine the length of a mathematical pendulum that makes one complete oscillation in 2 s

Determine the length of a mathematical pendulum that makes one complete oscillation in 2 s, if the acceleration due to gravity is 9.81 m / s2 How many times must the length of the pendulum be changed in order for its frequency to double?

T1 = 2 s.

g = 9.81 m / s2.

v2 = 2 * v1.

L1 -?

L2 / L1 -?

The time of one full swing is called the swing period T.

The period of the mathematical pendulum T is determined by the formula: T = 2 * п * √L / √g, where п is the number pi, L is the length of the pendulum, g is the acceleration of gravity.

T1 = 2 * п * √L1 / √g.

T12 = 4 * п^2 * L1 / g.

L1 = g * T1^2 / 4 * п^2.

L1 = 9.81 m / s2 * (2 s) ^ 2/4 * (3.14) ^ 2 = 0.994 m.

The frequency of oscillations v is the number of oscillations per unit of time, this value is inverse to the oscillation period T: v = 1 / T.

v1 = 1 / T1 = √g / 2 * P * √L1.

v2 = 1 / T2 = √g / 2 * P * √L2.

√g / 2 * P * √L2 = 2 * √g / 2 * P * √L1.

1 / √L2 = 2 / √L1.

1 / L2 = 4 / L1.

L2 = L1 / 4.

Answer: the mathematical pendulum has a length L1 = 0.994 m, its length must be reduced by 4 times: L2 = L1 / 4.