A body of mass m is suspended on a spiral spring with a stiffness of 100 N / m. if this system is taken

A body of mass m is suspended on a spiral spring with a stiffness of 100 N / m. if this system is taken out of equilibrium, then the body will perform 300 vibrations per minute. what is body weight?

Data: k is the stiffness of the coil spring used (k = 100 N / m); N is the number of perfect vibrations (N = 300 vibrations); t is the duration of vibrations of the suspended body (t = 1 min, which in the SI system is 60 s).

To find out the desired mass of the suspended body, consider the equality: 2 * Π * √ (mx / k) = T (period) = t / N, whence we express: mx / k = (t / (2 * Π * N)) ^ 2 and mx = k * (t / (2 * Π * N)) ^ 2.

Let’s perform the calculation: mx = k * (t / (2 * Π * N)) ^ 2 = 100 * (60 / (2 * 3.14 * 300)) ^ 2 ≈ 0.1 kg.

Answer: The mass of the body suspended from the spiral spring used should be 0.1 kg.