# The amplitude of the oscillations of the load on the spring is 5 cm, and the period is 2 s.

**The amplitude of the oscillations of the load on the spring is 5 cm, and the period is 2 s. How long does it take for the load to travel 40 cm?**

A = 5 cm = 0.05 m.

T = 2 s.

S = 40cm = 0.4m.

t -?

Path S is the length of the trajectory that the body describes as it moves. The period of oscillation T is called the time of one complete oscillation. The body from the equilibrium position for a time equal to the period travels the path A to the extreme equilibrium position, returns from the extreme position to the equilibrium position, passes the path A and so on to the second extreme position.

In one period T = 2 s, the body traverses the path S1 = 4 * A = 4 * 0.05 m = 0.2 m.

Let’s find the number of periods n: n = S / S1 = S / 4 * A = 0.4 m / 0.2 m = 2.

The time t, during which the body travels the path S, can be expressed by the formula: t = n * T.

t = 2 * 2 s = 4 s.

Answer: the body travels the specified path in time t = 4 s.