A body weighing 2 kg is suspended from a spring. If a body weighing 300 g is attached to it, then the spring will

A body weighing 2 kg is suspended from a spring. If a body weighing 300 g is attached to it, then the spring will stretch by another 2 cm. What will be the period of oscillation if the three-hundred-gram attachment is removed and the body weighing 2 kg is allowed to oscillate?

Let a body with a mass m₁ = 2 kg be suspended from a spring with a stiffness k. It creates an elongation of the spring x₁ meters. Spring force | Fprуп | = k ∙ х₁ will be balanced by the weight of the cargo Р₁ = m₁ ∙ g, where the coefficient g = 9.8 N / kg, that is, k ∙ х₁ = m₁ ∙ g. Then х₁ = m₁ ∙ g / k. If a body weighing m₂ = 300 g = 0.3 kg is attached to the load, then the spring will stretch additionally by x на = 2 cm = 0.02 m.In the second case, Fcont = Р₂ or k ∙ (x₁ + x₂) = (m₁ + m₂) ∙ g, then х₁ = – х₂ + (m₁ + m₂) ∙ g / k. We get, m₁ ∙ g / k = – x₂ + (m₁ + m₂) ∙ g / k, rigidity k = m₂ ∙ g / x₂.

To determine what the oscillation period will be if the three-hundred-gram add-on is removed and the body weighing 2 kg is allowed to oscillate, we will use the formula T = 2 π √ (m₁ / k), where the coefficient π ≈ 3.14. Let’s combine the formulas:

Т = 2 · π · √ (m₁ ∙ х₂ / (m₂ ∙ g)).

Substitute the values ​​of physical quantities in the calculation formula:

T = 2 · 3.14 · √ (2 kg ∙ 0.02 m / (0.3 kg ∙ 9.8 N / kg));

T = 2.15 s.

Answer: the oscillation period will be 2.15 seconds.