A ball weighing 0.2 kg, suspended on a spring, vibrates with a frequency of v = 5 Hz. Determine the coefficient of elasticity of the spring.

m = 0.2 kg.
v = 5 Hz.
k -?
The natural frequency of a spring pendulum is determined by the formula: v = √k / 2 * P * √m, where k is the stiffness of the spring, P is the number pi, m is the mass of the load.
k = 4 * P ^ 2 * m * v ^ 2.
k = 4 * (3.14) ^ 2 * 0.2 kg * (5 Hz) ^ 2 = 197 N / m.
Answer: spring rate k = 197 N / m.