A load suspended on a spring with a stiffness of 50 N / m vibrates with a frequency of 4 Hz

A load suspended on a spring with a stiffness of 50 N / m vibrates with a frequency of 4 Hz with what frequency will the same vibrate if it is suspended on a spring with a stiffness of 200 N / m?

k1 = 50 N / m.

v1 = 4 Hz.

k2 = 200 N / m.

v2 -?

The frequency of oscillations v is called the number of oscillations per unit of time, that is, for 1 s.

For a spring pendulum, the frequency of its natural vibrations v is determined by the formula: v = √k / 2 * п * √m, where k is the spring stiffness, п is the number pi, and m is the mass of the load.

v1 = √k1 / 2 * п * √m.

√m = √k1 / 2 * п * v1.

v2 = √k2 / 2 * п * √m = √k2 * 2 *п * v1 / 2 *п * √k1 = √k2 * v1 / √k1.

v2 = √200 N / m * 4 Hz / √50 N / m = 8 Hz.

Answer: the frequency of natural oscillations of the same weight on another spring will be v2 = 8 Hz.