Stretching a spring by x takes work. But the force of the spring is F = kx, growing the further it is stretched. It is not constant, so multiplying force by distance will not do.
Draw the F–x graph and the force is a straight line through the origin. The work is the area under that line, the area of a triangle. Base x, height kx, so the area is \dfrac{1}{2}kx^2. That is the energy U stored in the spring.
The \dfrac{1}{2} appears because the average of the force applied is \dfrac{1}{2}kx, which comes to the same thing. It grows uniformly from 0 to kx, so the average sits exactly halfway.
As the weight moves, the kinetic energy K = \dfrac{1}{2}mv^2 and the spring energy U = \dfrac{1}{2}kx^2 trade places. Only the dividing line inside the bar moves; the full length does not.
At the ends the speed is 0, so all of it is U; at the centre the extension is 0, so all of it is K. The v = 0 at the ends and a = 0 at the centre from the second lecture have become a bar.
The sum always equals its value at the ends: E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2 = \dfrac{1}{2}kA^2. K and U each reach 0 twice in one round trip, so watching those two alone makes the oscillation look twice as fast as it is.
Solve the energy equation for v. From \dfrac{1}{2}mv^2 = \dfrac{1}{2}kA^2 - \dfrac{1}{2}kx^2 comes v = \pm\omega\sqrt{A^2 - x^2}.
The time t has gone. Without knowing what moment it is, knowing where the weight is gives its speed. This is the most useful tool of the lot in examinations. The sign is \pm because the weight passes the same position going right and going left.
Plot v up the side and x across and the result is an ellipse. The circle that was turning in the first two lectures returns here as an ellipse. The point goes once round it while the weight makes one round trip.
Set two oscillations of different amplitude side by side. The period is unchanged, and the two weights reach their ends together. As isochronism in the first lecture said, the amplitude has no effect on the period.
The energy is another matter. Since E = \dfrac{1}{2}kA^2, halving the amplitude leaves a quarter. The bars stand in the ratio of four to one.
Doubling the amplitude leaves the round trip taking the same time, and yet the energy needed to get there is four times as great. Period against amplitude and energy against amplitude behave in completely different ways.