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 (0.66)A— that is, when its position is between −A and(−0.66)A, and when its position is between (0.66)A and A. What total percentage of the period does the mass lie in these regions?

54%

Explanation:

So, we have that the "magnitude of its displacement from equilibrium is greater than (0.66)A—''. Thus, the first step to take in answering this question is to write out the equation showing the displacement in simple harmonic motion which is = A cos w×t.

Therefore, we will have two instances t the displacement that is to say at a point 2π/w - a2 and the second point at a = a2.

Let us say that 2π/w = A, then, we have that a = A cos ^-1 (0.66)/2π. Also, we have that a2 = A/2 - A cos^- (0.66) / 2π.

The next thing to do is to calculate or determine the total length of of the required time. Thus, the total length is given as:

2a1 + ( A - 2a2) = 2A{ cos^-1 (0.66)}/ π.

Therefore, the total percentage of the period does the mass lie in these regions = 100 × {2a1 + ( A - 2a2) }/A = 2 { cos^-1 (0.66)}/ π × 100 = 54%.

Thus, the total percentage of the period does the mass lie in these regions = 54%.

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6