[color=#0000ff][i][color=#0000ff][i][color=#999999]This activity belongs to the GeoGebra book [url=https://www.geogebra.org/m/mes4bgft]The Domain of the Time[/url].[/color][/i][/color][/i][/color][br][br]Simple Harmonic Motion (SHM) can be seen as a projection of Uniform Circular Motion (UCM). Let's see how.[br][br]The following construction is designed so that the green point [color=#00ff00]MM[/color] follows a UCM around O, but always remains vertically aligned with the blue point [color=#0000ff]M[/color]. This implies that the radius of the circle must be equal to the amplitude [i]A[/i] of the SHM. The centripetal acceleration of this UCM is [b][color=#6aa84f]c[/color][/b] (green dashed vector), whose magnitude we have seen equals [color=#6aa84f][color=#ff3366][i]ω[/i][/color][/color][sup]2[/sup] [i]A[/i]. [br][br]To keep [color=#00ff00]MM[/color] always vertically aligned with [color=#0000ff]M[/color], the horizontal component of [color=#6aa84f][b][b]c[/b][/b][/color] must be precisely the acceleration [color=#6aa84f][b]a[/b][/color] of the SHM. This is because the triangle with [color=#6aa84f][b]a[/b][/color] as a leg and [color=#6aa84f][b][b]c[/b][/b][/color] as the hypotenuse must be similar to the triangle with [i]x[/i] as a leg and [i]A[/i] as the hypotenuse, as shown in the construction.[br][br]Due to this similarity of triangles, we have that |[color=#6aa84f][b]a[/b][/color]|/|[b][b][color=#6aa84f]c[/color][/b][/b]| = x/[i]A[/i], meaning |[color=#6aa84f][b]a[/b][/color]| = |[b][b][color=#6aa84f]c[/color][/b][/b]| [i]x[/i]/[i]A[/i] = [color=#6aa84f][color=#ff3366][i]ω[/i][/color][/color][sup]2[/sup] [i]x[/i]. [br][br]But we also knew that |[color=#6aa84f][b]a[/b][/color]| = [i]k[/i]/[i]m[/i] [i]x[/i], so it must hold that [color=#6aa84f][color=#ff3366][i]ω[/i][/color][/color][sup]2[/sup] = [i]k[/i]/[i]m[/i], meaning the angular velocity of the UCM must be [color=#ff3366][i]ω[/i][/color] = [math]\sqrt{k\slash m}[/math] and the period must be [i]T[/i] = 2π/[color=#6aa84f][color=#ff3366][i]ω[/i][/color][/color] = 2π[math]\sqrt{m\slash k}[/math]. Since every time [color=#00ff00]MM[/color] makes one full revolution, [color=#0000ff]M[/color] completes one full oscillation, the period of both motions must be the same.[br][br]We then deduce that [b][color=#cc0000]the period of the SHM does not depend on the amplitude [i]A[/i][/color][/b]; it only depends on the mass [i]m[/i] and the elasticity [i]k[/i]. For a spring with elasticity [i]k[/i], the mass [i]m[/i] will always take the same amount of time to complete a full oscillation. This property is known as [i]isochronism[/i].

[b]SCRIPT FOR SLIDER anima[/b][br][br][color=#cc0000][color=#cc0000]# Calculate the elapsed seconds dt; add one second if t1(1) < tt[/color][/color][br][color=#999999]SetValue(tt, t1(1))[br]SetValue(t1, First(GetTime(), 3))[br]SetValue(dt, (t1(1) < tt) + (t1(1) − tt)/1000)[/color][br][br][color=#cc0000]# Move M and MM[/color][br][color=#0000ff]SetValue(MM, O + (r; t ω))[/color][color=#999999][br][color=#999999]SetValue[/color](aux, v)[br][color=#999999]SetValue[/color](v, v + dt a)[br][color=#999999]SetValue[/color](M, M + dt v)[/color][br][br][color=#cc0000]# Record the period time and the number of complete oscillations[/color][br][color=#999999][color=#999999]SetValue[/color](reg, If(x(aux) > 0 ∧ x(v) < 0, Append(t, reg), reg))[br][color=#999999]SetValue[/color](osci, If(x(aux) > 0 ∧ x(v) < 0, osci + 1, osci))[br][/color][color=#0000ff][br][br][br][br][br][color=#999999][color=#999999][color=#0000ff][color=#0000ff][color=#999999][color=#999999]Author of the activity and GeoGebra construction: [/color][/color][/color][color=#0000ff][color=#999999][color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url].[/color][/color][/color][/color][/color][/color][/color]