A body weighing 0.2 kg is suspended on a spring with a stiffness of 2000 nm. Determine the frequency

A body weighing 0.2 kg is suspended on a spring with a stiffness of 2000 nm. Determine the frequency of free vibrations of this body.

To find the desired vibration frequency of a given body, we will use the formula: ν = 1 / (2 * Π) * √ (k / m).

Variables: k – spring stiffness (k = 2000 N / m); m – body weight (m = 0.2 kg).

Calculation: ν = 1 / (2 * 3.14) * √ (2000 / 0.2) = 15.924 Hz.

Answer: Free vibrations of the taken body occur with a frequency of 15.924 Hz.



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