Quiz2 Spring-mass-smooth zzz-graphs - Microsoft Excel X File Hor Inst Pag For Dat Rev Viel Dev a ? - Q X E - 27- Paste Font Alignment Number S Styles Cells 2. Clipboard Editing F41 Y A B C D E F 27 28 mass 2-kg spring-300N/m spring friction=0 29 30 approx time = elapsed time + (x2-x1)/((v2+v1)/2) 31 elap + dt meter m/sec 32 0.2 33 0.054433 0.16 1.469694 34 0.077762 0.12 1.959592 35 0.096788 0.08 2.244994 36 0.114011 0.04 2.4 37 0.130508 0 2.44949 38 39 0.128255 theory quarter cycle time 40 41 42 43 1 4 > > Sheet1 Sheet2 Sheet3 ji Ready @ @ 100% Question 1 Velocity, v(x), is a function of position x Energy conservation 0.5*2*300*0.2*0.2 - 0.5*300*x*x = 0.5*2*v(x)*v(x) Redo for 10 time intervals, find the total elapsed time for 1/4 cycle. Spring-mass-friction Mass 2 kg Spring 300 N/m Kinetic friction coefficient 0.35 Initial compression 0.2 mzzz-graphs Microsoft Excel File Home Insert Page Layout Formulas Data Review View Developer 6 - 5 x - 11 . A A = Wrap Text General Paste S - % ' *68 : Conditional F 1 Format Cell Insert Delete Format Sort & Find & BIU . B . DA. = iF Merge & Center . Formatting . as Table ~ Styles 2 7 Filler ~ Select- Clipboard Font Alignment Humber Styles Cells Editing 122 f =0.5 300"H22"H22+0.35 2 9.8"H22 L M N A B C D K 18 mass 2-kg spring-300N/m spring friction-0.35 4.628 1.372 2.50096 1.05644 1.00872 0.74088 19 3.5574 1.7496 20 x= 0.2 m compress initial 0 m/s 2.15127869 1.581442 1.004351 21 0.5 300*0.2 0.2 - 0.35 2'9.8'0.2= KE passing neutral & elongate 22 6 - 1.372 = 4.628 0.05 0.718 0.05 0.718 0.05 23 v = 2.1513 m/s passing neutral & elongate 0.1 2.186 0.1 2.186 0.1 24 4.628 = 0.5 300*x"x + 0.35'2*9.8*x when x = new amplitude 0.15 4.404 0.15 4.404 0.15 25 x = 0.154 meter, use graphical intersection/quadratic formula/Excel solver 0.2 7.372 0.2 7.372 0.2 0+ 26 V= 0 0.1 0.2 27 1.05 4.628 0.05 2.5009 0.05 1.00872 28 0.5 300*0.154 0.154 - 0.35 2 9.8 0.154=KE passing neutral & compress 0.1 4.628 0.1 2.50096 0.1 1.00872 29 3.5574-1.05644 = 2.50096 ).15 4.628 0.15 2.50096 0.25 30 v= 1.5814 m/s passing neutral& compress 0.2 4.628 0.2 2.50096 31 2.50096 = 0.5 300"x"x +0.35*2'9.8"x when x = new amplitude 32 x = 0.108 meter 0.2 33 v=0 approx time = elapsed time + (x2-x1)/((v2+v1)/2) 34 elap + dt meter m/sec 0.15 35 0.2 0 compress amplitude 36 0.5 300*0.108 0.108-0.35 2 9.8*0.108= KE passing neutral & elongate 0.185934086 2.1513 1st neutral 0.1 37 1.7496-0.74088 = 1.00872 0.329103333 0.154 0 elongate amplitude 38 v= 1.0043 m/s passing neutral& elongate 0.523867466 1.5814 2nd neutral 39 1.00872= 0.5 300"x"x +0.35 2'9.8"x when x = new amplitude 0.660455299 0.108 0 compress amplitude |0.05 40 x = 0.0622 meter 0.875530476 0 1.0043 3rd neutral 41 v=0 0.999397847 0.0622 0 elongate amplitude 42 02 0.4 0.6 0.8 1 . M Sheetl Sheetz Sheet4 Sheet30 Ready 100% Question 2a When the initial compression was 0.15 m, Complete the follow table using TI calculator, On-line calculator, Excel graphical intersection method, or Excel solver to do the quadratic formula in the computation of the energy conservation constraint condition Position meters Speed m/s Elapsed time Initial amplitude 0.15 m 0 m/s 0 compression mode First neutral position 0 ? ? ?? New amplitude ?? 0 m/s ?? elongation mode Second neural position 10 ?? ?? New amplitude 0 m/s ? ? compression mode Third neutral position 10 ? ? ?? New amplitude ? ? 0 m/s ? ? elongation mode Question 2b Use Excel graphing tool to show the non-simple harmonic motion oscillation with friction in a graph of position y-axis versus time x-axis