(*^ ::[ fontset = title, "Helv", 22, L0, center, nohscroll, bold; fontset = subtitle, "Helv", 18, L0, center, nohscroll, bold; fontset = subsubtitle, "Helv", 12, L0, center, nohscroll, bold; fontset = section, "Helv", 12, L0, bold, grayBox; fontset = subsection, "Helv", 12, L0, bold, blackBox; fontset = subsubsection, "Helv", 10, L0, bold, whiteBox; fontset = text, "Helv", 12, L0; fontset = smalltext, "Helv", 10, L0; fontset = input, "Courier", 12, L0, nowordwrap; fontset = output, "Courier", 12, L0, nowordwrap; fontset = message, "Courier", 10, L0, nowordwrap, R65280; fontset = print, "Courier", 10, L0, nowordwrap; fontset = info, "Courier", 10, L0, nowordwrap; fontset = postscript, "Courier", 10, L0, nowordwrap; fontset = name, "Helv", 10, L0, nohscroll, italic, B65280; fontset = header, "Helv", 18, L0, nohscroll, bold; fontset = footer, "Helv", 18, L0, center, nohscroll, bold; fontset = help, "Helv", 10, L0, nohscroll; fontset = clipboard, "Helv", 12, L0, nohscroll; fontset = completions, "Helv", 12, L0, nowordwrap, nohscroll; fontset = network, "Courier", 10, L0, nowordwrap, nohscroll; fontset = graphlabel, "Courier", 10, L0, nowordwrap, nohscroll; fontset = special1, "Helv", 12, L0, nowordwrap, nohscroll; fontset = special2, "Helv", 12, L0, center, nowordwrap, nohscroll; fontset = special3, "Helv", 12, L0, right, nowordwrap, nohscroll; fontset = special4, "Helv", 12, L0, nowordwrap, nohscroll; fontset = special5, "Helv", 12, L0, nowordwrap, nohscroll; fontset = Left Header, "Helv", 12, L0, nowordwrap, nohscroll; fontset = Left Footer, "Helv", 12, L0, nowordwrap, nohscroll;] :[font = subsection; inactive; ] Example 7. :[font = text; inactive; ] A mass of 5 Kg is attached to a spring hanging from the ceiling, thereby stretching the spring 0.5 m on coming to rest at equilibrium. The mass is then pulled down 0.1 m below the equilibrium point and given an upward velocity of 0.1 m/sec. When will the mass first reach its minimum height after being set in motion? :[font = text; inactive; ] The equation x[t] of this simple harmonic motion and its graph (over a finite interval)are given by: :[font = input; startGroup; ] Plot[1/10 Cos[7/5 Sqrt[10] t]-1/14 Sqrt[10]/10 Sin[7/5 Sqrt[10] t], {t,-Pi,Pi}] :[font = postscript; inactive; output; BITMAP; PostScript; pictureLeft = 100; pictureTop = 0; pictureWidth = 300; pictureHeight = 185; ] %! %%Creator: Mathematica %%AspectRatio: 0.61803 MathPictureStart /Courier findfont 10 scalefont setfont % Scaling calculations 0.5 0.151576 0.309017 2.87071 [ [(-3)] 0.04527 0.30902 0 2 Msboxa [(-2)] 0.19685 0.30902 0 2 Msboxa [(-1)] 0.34842 0.30902 0 2 Msboxa [(1)] 0.65158 0.30902 0 2 Msboxa [(2)] 0.80315 0.30902 0 2 Msboxa [(3)] 0.95473 0.30902 0 2 Msboxa [(-0.1)] 0.4875 0.02195 1 0 Msboxa [(-0.05)] 0.4875 0.16548 1 0 Msboxa [(0.05)] 0.4875 0.45255 1 0 Msboxa [(0.1)] 0.4875 0.59609 1 0 Msboxa [ -0.001 -0.001 0 0 ] [ 1.001 0.61903 0 0 ] ] 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As the graph suggests, x[t] is maximum at the second stationary point (t>0). Clearly without the graph, the problem would have been very difficult to do. Now to find the minimum height, the following steps may be followed: :[font = text; inactive; ] Let T=7/5 Sqrt[10] t. :[font = input; startGroup; ] x[T]= 1/10 Cos[T] -1/14 Sqrt[10]/10 Sin[T] :[font = output; inactive; formatted; output; endGroup; ] Cos[T]/10 - Sin[T]/(7*40^(1/2)) ;[o] Cos[T] Sin[T] ------ - ---------- 10 7 Sqrt[40] :[font = input; startGroup; ] x'[T]=D[x[T],T] :[font = output; inactive; formatted; output; endGroup; ] -Cos[T]/(7*40^(1/2)) - Sin[T]/10 ;[o] -Cos[T] Sin[T] ---------- - ------ 7 Sqrt[40] 10 :[font = input; startGroup; ] Solve[x'[T]==0] :[font = output; inactive; formatted; output; endGroup; ] {{Cos[T] -> -7*(2/5)^(1/2)*Sin[T]}} ;[o] 2 {{Cos[T] -> -7 Sqrt[-] Sin[T]}} 5 :[font = text; inactive; ] This means that Tan[T]= -1/7 Srqt[5/2]]. :[font = input; startGroup; ] N[ArcTan[-1/7 Sqrt[5/2]]] :[font = output; inactive; formatted; output; endGroup; ] -0.2221490035351628 ;[o] -0.222149 :[font = text; inactive; ] In fact, T=-0.222149 + n[Pi], n=...-2, -1, 0, 1, 2, .... Since T is positive, then n is at least equal to 1. Since, the minimum height occurs at the second stationary point then n=2. Hence, :[font = input; startGroup; ] T=N[2 Pi-0.222149] :[font = output; inactive; formatted; output; endGroup; ] 6.061036307179587 ;[o] 6.06104 :[font = text; inactive; ] Now we solve for t: :[font = input; startGroup; ] t=N[5*6.06104/Sqrt[490]] :[font = output; inactive; formatted; output; endGroup; ] 1.369049384956211 ;[o] 1.36905 ^*)