A horizontal spring is attached to a wall at one end and a mass at the other. The mass rests on a frictionless surface. You pull the mass, stretching the spring beyond the equilibrium position a distance A, and release it from rest. The mass then begins to oscillate in simple harmonic motion with amplitude A. During one period, the mass spends part of the time in regions where the magnitude of its displacement from equilibrium is greater than (0.30)A— that is, when its position is between −A and (−0.30)A, and when its position is between (0.30)A and A. What total percentage of the period does the mass lie in these regions?

 Δt'/ T% = 90.3%

Explanation:

Simple harmonic movement is described by the expression

         x = A cos (wt)

we find the time for the two points of motion

x = - 0.3 A

        -0.3 A = A cos (w t₁)

         w t₁ = cos -1 (-0.3)

remember that angles are in radians

        w t₁ = 1.875 rad

x = 0.3 A

        0.3 A = A cos w t₂

        w t₂ = cos -1 (0.3)

         w t₂ = 1,266 rad

Now let's calculate the time of a complete period

x= -A

        w t₃ = cos⁻¹ (-1)

        w t₃ = π rad

this angle for the forward movement and the same time for the return movement in the oscillation to the same point, which is the definition of period

         T = 2 t₃

         T = 2π / w     s

now we can calculate the fraction of time in the given time interval

        Δt / T = (t₁ -t₂) / T

        Δt / T = (1,875 - 1,266) / 2pi

        Δt / T = 0.0969

This is the fraction for when the mass is from 0 to 0.3, for regions of oscillation of greater amplitude the fraction is

         Δt'/ T = 1 - 0.0969

         Δt '/ T = 0.903

         Δt'/ T% = 90.3%

i believe the answer is a

i already answered it by 34/s