What is the period of oscillation of a weight of 0.2 kg on a spring with a stiffness of 0.05 N / m?
| February 1, 2021 | education
These tasks: m (weight of the load on the formed spring pendulum) = 0.2 kg; k (spring rate) = 0.05 N / m.
The oscillation period of the load on the formed spring pendulum is calculated by the formula: Т = 2Π * √ (m / k).
Let’s do the calculation: T = 2 * 3.14 * √ (0.2 / 0.05) = 12.56 s.
Answer: The oscillation period of the load is 12.56 s.
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