The amplitude of the vibrations of the load on the spring is 10 cm, the frequency is 5 Hz. What path will the body take in 20s?

Let us find the period of oscillation of the load, this is the reciprocal of the oscillation frequency:

T = 1 / v.

T = 1/5 = 0.2 (s).

This means that the load makes one complete vibration in 0.2 s.

Since the amplitude, according to the condition of the problem, is 10 cm, and this is the maximum deviation from the equilibrium point, the path for one complete oscillation performed by the load will be:

10 * 4 = 40 cm.

Let us find how many complete oscillations the load will make in the required time:

20 / 0.2 = 100 (times).

Accordingly, the path will be covered in 20 seconds:

40 * 100 = 4000 (cm) = 40 (m).

Answer: 40 m.