A spring that can be stretched by x = 5.0cm under the action of a force F. If a load weighing m = 0.25kg

A spring that can be stretched by x = 5.0cm under the action of a force F. If a load weighing m = 0.25kg is suspended from this spring, then the period of vertical oscillations of the load will be equal to T = 1.2c. Find I. Disregard the mass of the spring.

Data: Δx (spring deformation) = 5 cm (in SI Δx = 0.05 m); m (weight of the suspended load) = 0.25 kg; T (period of vertical oscillations) = 1.2 s; the mass of the spring is not included.

1) Determine the stiffness of the spring: T = 2 * Π * √ (m / k); √ (m / k) = T / (2 * Π); m / k = T ^ 2 / (4 * Π ^ 2); k = m * 4 * Π ^ 2 / T2 = 0.25 * 4 * 3.14 ^ 2 / 1.22 = 6.85 N / m.

2) Calculate the force applied to the spring: F = k * Δx = 6.85 * 0.05 ≈ 0.34 N.

Answer: The effective force is 0.34 N.