A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionless surface below. Suppose you pull the mass so that the spring is stretched out (initial state) and then you release it, so that the mass starts moving towards the spring is unstretched position (final state). The impulse imparted on the spring-mass system by the force that the wall exerts on the spring is zero, since the wall does not move during this process. Required:
What total percentage of the period does the mass lie in these regions?

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

well, you would divide 135 by 33 to get your answer. then that is how many seconds it would take. boardman should fly about 132 m or 4.090909090909091 seconds.