(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 18067, 490]*) (*NotebookOutlinePosition[ 18961, 520]*) (* CellTagsIndexPosition[ 18917, 516]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Period, Frequency, Amplitude", "Subsection"], Cell[TextData[{ StyleBox["Things that Repeat", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\nToday we will discuss phenomena that ", StyleBox["repeat in time", FontSlant->"Italic"], ". There are some terms and models that apply to any system -- circular \ motion, stock market cycles, the displacement of your vocal chords while \ singing a C note -- which may be said to ", StyleBox["repeat in time", FontSlant->"Italic"], ". Name another system which repeats in time.\n\n", StyleBox["Period", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\nAny system that repeats in time in a predictable fashion can be \ characterized by the amount of time between repetitions, the ", StyleBox["period", FontSlant->"Italic"], ".\n\n", StyleBox[" Period ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox["T", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic", Background->GrayLevel[0.666667]], StyleBox[ " is a Length of Time between repetitions. It is usually expressed in \ Seconds. ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], "\n\n", StyleBox["Frequency", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\nEquivalently, you can say how ", StyleBox["many ", FontSlant->"Italic"], "(a number) fit in a certain amount of time -- the ", StyleBox["frequency", FontSlant->"Italic"], ".\n\n", StyleBox[" Frequency ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox["f", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic", Background->GrayLevel[0.666667]], StyleBox[" characterizes ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], Cell[BoxData[ \(TraditionalForm\`\(Repetitions\ of\ Cycle\)\/\(Amount\ of\ Time\)\)], Background->GrayLevel[0.666667]], StyleBox[". It is usually expressed in 1/s, or Hertz (Hz). ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], "\n\nIf a system takes 0.1 seconds to repeat, it must repeat 10 times a \ second. In general\n", StyleBox["\t", Background->GrayLevel[0.900008]], StyleBox[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[\(f = 1\/T\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Frequency vs. Period equation", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], "\n\t\nYou don't have to know anything else about the system... once you \ know its period, you always know its frequency. Once you know its frequency, \ you always know its period.\n\nEx. A turntable doing 33 rpm rotates at a \ frequency of\n\t", Cell[BoxData[ \(TraditionalForm \`33\ rev\/min\[CenterDot]\(1\ min\)\/\(60\ s\) = \(0.55\ rev\/s = 0.55\ Hz\)\)]], "\nWhat is the period of the turntable?\nNow the DJ puts the mix at 78 rpm, \ yo. Is the frequency higher or lower?\nWhat is the frequency (in Hz)?\nIs \ the period higher or lower?\nWhat is the period (in seconds)?\n\nPlease note: \ a Hertz is just a convenient name for the unit \"inverse seconds\", ", Cell[BoxData[ \(TraditionalForm\`1\/s\)]], ". There may be a \"number\" part on top -- for instance, \"", Cell[BoxData[ \(TraditionalForm\`revs\/sec\)]], "\" or \"", Cell[BoxData[ \(TraditionalForm\`\(#\ of\ waves\)\/sec\)]], "\". As long as that number has no ", StyleBox["mks", FontSlant->"Italic"], " units the whole quantity is equivalent to ", Cell[BoxData[ \(TraditionalForm\`1\/s\)]], ", or Hertz." }], "Text"], Cell[TextData[{ StyleBox["Amplitude", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\nWe can average a repeating phenomenon over time. The ups will generally \ cancel the downs, and we can describe the behavior as an oscillation around \ an average, or equilibrium, position. The maximum excursion from equilibrium \ is called the amplitude.\n\nFor the stock market, the amplitude may be in \ total market capitalization.\nFor an alternating electrical system, the \ amplitude may be voltage (V).\n\nFor the common systems remember that\n", StyleBox[" For Spring Problems, Amplitude is a distance (meters). ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox["\n", Background->GrayLevel[0.666667]], StyleBox[" For Pendulum Problems, Amplitude is an angle (degrees). ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], "\n" }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Harmonic Motion: Spring", "Subsection"], Cell[TextData[{ "The law for a spring (Hooke's Law) states that a spring extended from \ equilibrium imposes a restoring force equal to the extension:\n", StyleBox["\t ", Background->GrayLevel[0.900008]], Cell[BoxData[ FormBox[ FrameBox[\(F = \ \(-k\)\[VeryThinSpace]x\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.900008]], StyleBox[" ", Background->GrayLevel[0.900008]], StyleBox["Spring Force Law", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.900008]], StyleBox[" ", Background->GrayLevel[0.900008]], "\n\nIf we stretch the spring to the right, there is a force, proportional \ and opposite to the extension, pulling it leftwards to equilibrium. If we \ release it, it will zip back to the center -- but it will overshoot, as it \ now has momentum leftwards. The momentum will carrry it to a point on the \ left, opposite its initial release point, where it will pause (velocity \ zero). It will then zip rightwards, back through the equilibrium. As it \ passes through the equilibrium point the acceleration is zero, but now its \ rightward momentum carries it through to the point at which it was released. \ This will repeat indefinitely for a high quality spring, with no drag or \ dissipation.\n\nPosition is left max\t\t\tPosition is zero\t\t\t\tPosition is \ right max\nVelocity is zero \t\t\tVelocity is max (l/r)\t\t\t Velocity \ is zero\nAccel is max, to right\t\t\tAccel is zero\t\t\t\t Accel is max, to \ left\n|------------------------------------------------------|----------------\ -----------------------------------------------|\n\n", StyleBox[ " Acceleration is always Opposite and Proportional to Position. \n \ Velocity is large when Position & Acceleration are small. \n Velocity \ is small when Position & Acceleration are large. ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]] }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Harmonic Motion: Pendulum", "Subsection"], Cell[TextData[{ "For a pendulum near equilibrium,\n", StyleBox["\t", Background->GrayLevel[0.900008]], StyleBox[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ \(F = \ \(-\ m\)\ g\ sin\ \[Theta]\ \[TildeEqual] \ \(-m\)\ g\ \ \[Theta]\ \), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Pendulum Force Law", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], "\n\nIf we tip the pendulum to the right, there is a force, proportional \ and opposite to the extension, pulling it leftwards to equilibrium. If we \ release it, it will zip back to the center -- but it will overshoot, as it \ now has momentum leftwards. The momentum will carrry it to a point on the \ left, opposite its initial release point, where it will pause (velocity \ zero). It will then zip rightwards, back through the equilibrium. As it \ passes through the equilibrium point the acceleration is zero, but now its \ rightward momentum carries it through to the point at which it was released. \ This will repeat indefinitely for a high quality pendulum, with no drag or \ dissipation.\n\nAngle is left max\t\t\tAngle is zero\t\t\t\t Angle is right \ max\nVelocity is zero \t\t\tVelocity is max (l/r)\t\t\t Velocity is \ zero\nAccel is max, to right\t\t\tAccel is zero\t\t\t\t Accel is max, to left\ \n|------------------------------------------------------|--------------------\ -------------------------------------------|\n\n", StyleBox[ " Acceleration is always Opposite and Proportional to Angle. \n \ Velocity is large when Angle & Acceleration are small. \n Velocity is \ small when Angle & Acceleration are large. ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]] }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Natural Frequency", "Subsection"], Cell[TextData[{ "The force laws for spring motion and pendulum motion are suspiciously \ similar: ", Cell[BoxData[ \(TraditionalForm\`F\ = \ Junk\ \[CenterDot]\ Amplitude\)]], ". This is the simplest model we have in physics for a repeating system, \ and it is called a Simple Harmonic Oscillator.\n", StyleBox["\t", FontFamily->"Trebuchet MS"], StyleBox[ "(Public service announcement for the calculus-hep: recall that \ acceleration is the second derivative of position. If \n\t\t", FontFamily->"Trebuchet MS", FontSize->9], Cell[BoxData[ \(TraditionalForm\`F = \(-k\)\ x\)], FontFamily->"Trebuchet MS", FontSize->9], StyleBox[", then\n\t\t", FontFamily->"Trebuchet MS", FontSize->9], Cell[BoxData[ \(TraditionalForm\`m\ a\ = \ \(m\ x\^\[DoublePrime] = \ \(-k\)\ x\)\)], FontFamily->"Trebuchet MS", FontSize->9], StyleBox[", or\n\t\t", FontFamily->"Trebuchet MS", FontSize->9], Cell[BoxData[ \(TraditionalForm\`x\^\[DoublePrime] = \(-\(k\/m\)\) x\)], FontFamily->"Trebuchet MS", FontSize->9], StyleBox["\n\tThe simplest repeating system: the ", FontFamily->"Trebuchet MS", FontSize->9], StyleBox["second derivative", FontFamily->"Trebuchet MS", FontSize->9, FontSlant->"Italic"], StyleBox[ " of the excursion is proportional and opposite to the excursion itself. \ PSA over).", FontFamily->"Trebuchet MS", FontSize->9], "\n\nThe spring system has a solution in terms of the cosine of the \ amplitude:\n\t", Cell[BoxData[ \(TraditionalForm\`x\ = \ A\ \(cos(2\ \[Pi]\ \(f\&\ \)\ t)\)\)]], "\nThis is a repeating function of time, with amplitude ", Cell[BoxData[ \(TraditionalForm\`A\)]], " and frequency ", Cell[BoxData[ \(TraditionalForm\`f\)]], ". The frequency of the oscillation stems directly from the force law; it \ is the frequency of the system if disturbed and then left to cope. That ", StyleBox["natural frequency", FontSlant->"Italic"], " is given by\n", StyleBox["\t ", Background->GrayLevel[0.666667]], Cell[BoxData[ \(TraditionalForm\`f\ = \ \(1\/\(2 \[Pi]\)\) \@\(k\/m\)\)], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], StyleBox["Natural Frequency for Spring Force Law ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], StyleBox["\n\n\t ", Background->GrayLevel[0.666667]], Cell[BoxData[ \(TraditionalForm\`f\ = \ \(1\/\(2 \[Pi]\)\) \@\(\(g\&\ \)\/L\)\)], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], StyleBox["Natural Frequency for Pendulum Force Law ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], "\n\nNotice that the frequency does NOT depend on the amplitude.\n", StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Natural Frequency for Simple Harmonic Motion is ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], StyleBox["independent", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" of Amplitude. ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]] }], "Text"], Cell["\<\ Humans have a leg length of about 1 meter, and a stride of about 0.5 meter. What is the natural frequency of the human stride? What is the period of the human stride? What is the amplitude of the human stride? How does the natural frequency of a short person compare to that of a tall \ person? Adults tend to get carsick from frequencies below one Hz. Assume that we are \ comfortable with oscillations at or above our natural walking frequency but \ get carsick for frequencies below our natural walking frequency. Does a child get carsick more or less easily?\ \>", "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Kinetic and Potential Energy", "Subsection", Cell[TextData[{ "The law for a spring (Hooke's Law) states that a spring extended from \ equilibrium imposes a restoring force equal to the extension:\n", StyleBox["\t ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[\(PE = \ \(1\/2\) k\ x\^2\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Spring Potential Energy", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" \n\t ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[\(KE = \ \(1\/2\) m\ v\^2\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Spring Kinetic Energy", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], "\n\nIf we stretch the spring to the right, ", StyleBox["we", FontSlant->"Italic"], " do work ", StyleBox["on", FontSlant->"Italic"], " the spring, therefore storing potential energy in the spring.\nAs the \ spring relaxes back to equilibrium that potential energy becomes kinetic \ energy.\nThe kinetic energy stretches the spring back to the left, storing \ kinetic energy back in as potential energy.\n\nPosition is left max\t\t\t\ Position is zero\t\t\t\tPosition is right max\nVelocity is zero \t\t\t\ Velocity is max (l/r)\t\t\t Velocity is zero\nAccel is max, to right\t\t\ \tAccel is zero\t\t\t\t Accel is max, to left\nPE is maximum\t\t\tPE is zero\t\ \t\t\t PE is maximum\nKE is zero\t\t\t\tKE is maximum\t\t\t\t KE is \ zero\n|------------------------------------------------------|----------------\ -----------------------------------------------|\n\n", StyleBox[ " Acceleration is always Opposite and Proportional to Position. \n \ Velocity & KE are large when Position, Acceleration, & PE are small. \n \ Velocity & KE are small when Position, Acceleration, & PE are large. ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", FontFamily->"Bangkok", Background->GrayLevel[0.666667]] }], "Text"] }, Open ]] }, FrontEndVersion->"Microsoft Windows 3.0", ScreenRectangle->{{0, 1600}, {0, 1113}}, WindowSize->{897, 843}, WindowMargins->{{15, Automatic}, {Automatic, 52}}, PrintingCopies->1, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PrintingMargins"->{{36, 36}, {36, 36}}, "PrintCellBrackets"->False, "PrintRegistrationMarks"->False, "PrintMultipleHorizontalPages"->False}, Magnification->1.5 ] (*********************************************************************** Cached data follows. If you edit this Notebook file directly, not using Mathematica, you must remove the line containing CacheID at the top of the file. The cache data will then be recreated when you save this file from within Mathematica. ***********************************************************************) (*CellTagsOutline CellTagsIndex->{} *) (*CellTagsIndex CellTagsIndex->{} *) (*NotebookFileOutline Notebook[{ Cell[CellGroupData[{ Cell[1731, 51, 50, 0, 86, "Subsection"], Cell[1784, 53, 4036, 116, 867, "Text"], Cell[5823, 171, 977, 22, 314, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[6837, 198, 45, 0, 86, "Subsection"], Cell[6885, 200, 2119, 42, 481, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[9041, 247, 47, 0, 86, "Subsection"], Cell[9091, 249, 2129, 45, 459, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[11257, 299, 39, 0, 86, "Subsection"], Cell[11299, 301, 3646, 107, 537, "Text"], Cell[14948, 410, 591, 13, 261, "Text"] }, Open ]], Cell[CellGroupData[{ Cell[15576, 428, 50, 0, 86, "Subsection"], Cell[15629, 430, 2422, 57, 504, "Text"] }, Open ]] } ] *) (*********************************************************************** End of Mathematica Notebook file. ***********************************************************************)