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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[ 21268, 596]*) (*NotebookOutlinePosition[ 22156, 625]*) (* CellTagsIndexPosition[ 22112, 621]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Period, Frequency, Amplitude", "Section", PageWidth->PaperWidth], Cell[TextData[{ StyleBox["Things that Repeat", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\n", "Today 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." }], "Text"], Cell[TextData[{ StyleBox["Period", FontFamily->"Trebuchet MS", FontSize->12, FontWeight->"Bold"], "\nAny system that repeats in time in a predictable fashion can be \ characterized by the ", StyleBox["period", FontSlant->"Italic"], " - the amount of time between repetitions.\n", StyleBox["Period (", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox["T ", FontFamily->"Bangkok", FontWeight->"Bold", FontSlant->"Italic", Background->GrayLevel[0.666667]], StyleBox[")", FontFamily->"Times New Roman", FontWeight->"Bold", FontSlant->"Italic", Background->GrayLevel[0.666667]], StyleBox[" is the Length of Time between repetitions. It is expressed in \ Seconds. ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox["\n", FontSize->10], 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", 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 expressed in 1/s, or Hertz (Hz). ", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], "\n\nOnce you know the period of a system, you know its frequency, from the \ equation\n", StyleBox[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ FrameBox[\(f = 1\/T\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], 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]], "\nIf a system takes 0.1 seconds to repeat, it must repeat 10 times a \ second - 10 Hz.\n\nEx. A 33 rpm turntable rotates at a frequency of ", Cell[BoxData[ \(TraditionalForm\`33\ rev\/min\[CenterDot]\(1\ min\)\/\(60\ s\) = \ \(0.55\ rev\/s = 0.55\ Hz\)\)]], ". What is the period of the turntable? Now the DJ puts the mix at 78 rpm. \ Is the frequency higher or lower than before? Is the period higher or lower \ than before?\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", PageWidth->PaperWidth], 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. For 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]], "\nFor an alternating current electrical system, the amplitude is a current \ (", Cell[BoxData[ \(TraditionalForm\`I\_max\)]], ") -- the current swings from ", Cell[BoxData[ \(TraditionalForm\`\(+I\_max\)\)]], " to ", Cell[BoxData[ \(TraditionalForm\`\(-I\_max\)\)]], " about an equilibrium value of ", Cell[BoxData[ FormBox[ RowBox[{"I", "=", RowBox[{"0", StyleBox["A", FontSlant->"Plain"]}]}], TraditionalForm]]], ". Notice that the average value is 0A but the RMS value (the value used \ in Ohm's law, etc.) is ", Cell[BoxData[ \(TraditionalForm\`I\_rms = \(\@2\) I\_max\)]], "." }], "Text", PageWidth->PaperWidth], Cell[CellGroupData[{ Cell["Harmonic Motion: Spring", "Subsection", PageWidth->PaperWidth], 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[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ FrameBox[\(F = \ \(-\(k\&\ \)\)\[VeryThinSpace]x\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]], Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], StyleBox["Spring Force Law", FontFamily->"Bangkok", FontWeight->"Bold", Background->GrayLevel[0.666667]], StyleBox[" ", Background->GrayLevel[0.666667]], "\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 (one 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", PageWidth->PaperWidth] }, Open ]], Cell[CellGroupData[{ Cell["Harmonic Motion: Pendulum", "Subsection", PageWidth->PaperWidth], Cell[TextData[{ "For a pendulum near equilibrium,\n", StyleBox[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ RowBox[{ FrameBox[\(F = \ \(-\ m\)\ g\ sin\ \[Theta]\ \[TildeEqual] \ \(-m\)\ g\ \ \(\ \[Theta]\&\ \)\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], " "}], 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]], "\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\t Angle is \ right max\nVelocity is zero \t\t\tVelocity is max (l/r)\t\t\t\t \ Velocity is zero\nAccel is max, to right\t\t\tAccel is zero\t\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", PageWidth->PaperWidth] }, Open ]], Cell[CellGroupData[{ Cell["Natural Frequency", "Subsection", PageWidth->PaperWidth], 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. The 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 Kim's Equation:\n", StyleBox[" ", 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]], "\t\"Kim frequently drops her two pies\"", StyleBox["\n ", 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]], "\t\"Gil frequently drops his two pies\"\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. \n", FontFamily->"Bangkok", Background->GrayLevel[0.666667]], "Notice that the frequency does NOT depend on the amplitude.\n\n\t\"The \ more things change, the more they stay the same\"" }], "Text", PageWidth->PaperWidth], 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? The period? The amplitude? \ 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 \ oscillations at or above our natural walking frequency are comfortable but \ that frequencies below our natural walking frequency induce carsickness. \ Should children get carsick more or less easily?\ \>", "Text", PageWidth->PaperWidth] }, Open ]], Cell[CellGroupData[{ Cell["Kinetic and Potential Energy", "Subsection", PageWidth->PaperWidth], 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[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ FrameBox[\(PE\[VeryThinSpace] = \[VeryThinSpace]\(1\/2\) k\ x\^2\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], 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[" ", Background->GrayLevel[0.666667]], Cell[BoxData[ FormBox[ FrameBox[ FrameBox[\(KE = \(1\/2\) m\[VeryThinSpace]v\^2\), BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], 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 an object compresses the spring, that makes the system mad; \ therefore it\n\tCompressing:\tIncreases PE; decreases KE; system does \ negative work on object.\nAs the spring relaxes back to equilibrium (string \ uncompresses to its default position), the system is happy:\n\tRelaxing:\t\ Decreases PE; increases KE; system does positive work on object.\t\nAs the \ system passes equilibrium, the object still has KE; thus, it overshoots, \ making the system mad again:\n\tStretching:\tIncreases PE; decreases KE; \ system does negative work on object.\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\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\nFor Simple Harmonic Motion,\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", PageWidth->PaperWidth], Cell["In-Class Compendium, Passages 11 and 12 (time: 18:30)", "Subsubsection", PageWidth->PaperWidth] }, Open ]] }, Open ]], Cell[CellGroupData[{ Cell["Waves", "Section", PageWidth->PaperWidth, PageBreakAbove->True], Cell["Transverse (string, water, light) vs. Longitudinal (sound); ", "Text", PageBreakAbove->Automatic], Cell[TextData[{ "Extremely Important Wave Formula:\n\t", Cell[BoxData[ FormBox[ FrameBox[ FrameBox[ FormBox[\(\[Lambda]\ f\ = \ v\), "TraditionalForm"], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], BoxMargins->{{0.2, 0.2}, {0.4, 0.4}}], TraditionalForm]]], "\t(\[Lambda], wave", StyleBox["length:", FontSlant->"Italic"], " meters; ", StyleBox["f", FontSlant->"Italic"], ", frequency: \"per seconds\"; gives ", StyleBox["v", FontSlant->"Italic"], ", meters per second.\nWhat is the wavelength of radio station FM 99.1? \ What frequency is the size of an atom? of a cell?\nA rule of thumb is that a \ speaker cannot reproduce sound whose wavelength is larger than its diameter. \ What is the frequency cutoff for a 10\" (0.25 m) speaker." }], "Text", PageWidth->PaperWidth, PageBreakAbove->Automatic] }, Open ]], Cell[CellGroupData[{ Cell["Interference and Resonance", "Section", PageWidth->PaperWidth], Cell[CellGroupData[{ Cell["Constructive and Destructive Interference", "Subsubsection", PageWidth->PaperWidth], Cell["\<\ Standing wave node / anti-node fundamental\ \>", "Text", PageWidth->PaperWidth] }, Open ]], Cell[CellGroupData[{ Cell["Resonant Wavelengths and Frequencies", "Subsubsection", PageWidth->PaperWidth], Cell[TextData[{ "Draw pictures -- not formulas.\nResonant Frequency:\n\t", Cell[BoxData[ \(TraditionalForm\`f\_n = n\ f\_1\)]], "\t\tThe ", StyleBox["n", FontSlant->"Italic"], "'th resonant frequency is simply ", StyleBox["n", FontSlant->"Italic"], " times the fundamental.\ngiving", "\t\t\t", Cell[BoxData[ \(TraditionalForm\`\[Lambda]\_n = \(2 L\)\/n\)]], "\n\t\t\t", Cell[BoxData[ \(TraditionalForm\`f\_n = n v\/\(2 L\)\)]], "\nThe velocity of a standing wave on a stretched string is \n\t", Cell[BoxData[ \(TraditionalForm\`v = \@\(\[Tau]\/\(M/L\)\)\)]], ",\tvelocity = ", Cell[BoxData[ \(TraditionalForm\`\@\(tension\/\(mass/length\)\)\)]], "\nEx. A stand-up bass has 1.5m strings which weigh 50 g, tensioned to 3000 \ N. What is its fundamental frequency? If the string is plucked at the \ midpoint, what will the frequency and wavelength of the harmonic be?" }], "Text", PageWidth->PaperWidth] }, Open ]], Cell["In-Class Compendium, Passages 4 and 5 (time: 18:00)", "Subsubsection", PageWidth->PaperWidth], Cell["If time -- In-Class Compendium, Passage 15 (time: 8:00)", \ "Subsubsection", PageWidth->PaperWidth] }, Open ]] }, FrontEndVersion->"4.1 for Microsoft Windows", ScreenRectangle->{{0, 1600}, {0, 1087}}, WindowSize->{1065, 815}, WindowMargins->{{2, Automatic}, {Automatic, 17}}, PrintingCopies->1, PrintingPageRange->{Automatic, Automatic}, PrintingOptions->{"PrintingMargins"->{{36, 36}, {36, 36}}, "PrintCellBrackets"->False, "PrintRegistrationMarks"->False, "PrintMultipleHorizontalPages"->False}, Magnification->1.5 ] (******************************************************************* Cached data follows. 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