Class Note for MATH 254 at UA
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Date Created: 02/06/15
MATH 254 Forced damped oscillators Example 1 Spring hanging from a hinge external Mg MgikzZibFt gravity spring friction force force WVWWVWW Ln V I W M bkzFtFocoswt 2 L273 13 q Vt Vocoswt a b kl Mg Ft zt 0 is equilibrium Example 2 equilibrium In 147105 Wm am Example 3 LRC Circuit 0000000000 N C 5 4 qt charge Lg 13 g4 Vt i applied voltage Spring sliding on a plane dQI dz MW 7kribEFt Example 4 Pendulum m i 7mg sinz 7 Mb ZFt For small I sinz N z mfg b In gm l force Solution We Will assume in all cases that b2 lt 4km and R2 lt 4i Solutions of homogeneous equations7 10 t and go 7i 7 122 if b2 e 2 Acos m 74 tBsm m TmQt A cosg07 6710008 iiitigy VA2B2 m 47212 7 B 7 SWquot m a 10 t in L713 71372 b qct 6 2L Acos LC 4L2 tBsm LC 4L2t 2 67ECCOSltN 7t7 gt7 CA2B2 spring LRC Circuit spring LRC Circuit In the case ofthe the displacement of the nonforced decays With oscillations 27f The time between oscillation maxima is and the amplitude drop from one 27f or i 7 52 L 7 32 m 4m2 LC E b2 k R2 1 1R24L2 i Which7 1f TmQ ltlt m and TLQ ltlt 7L0 1s 1 2w HI L 2 m2 72 max1mum to the next 1s 6 m b 4 or e KR 67 ore Note ECO HOastHoo i 40 Particular Solution I Will work out the case for the LRC circuiti As an exercise7 you do the case for the spring Let qpt A1 cos wt A2 sinwti Substitute in b L7Alw2 coswt 7 Agw2 sinwt R7A1w sinwt Agw cos wt A1 cos wt A2 sinwt V0 coswt Divide across by w convenient and match coef cients of cos wt7 sinwt coswt 7wLgt A1RA2 lV0 wC w 0 sinwt 7RA1 7 wLgt A2 0 Solve for A1 and Ag it is convenient to de ne R as the circuit resistanceo7 and Zw 4 7 wL2 R2 as the circuit impedance To solve for Al7 multiply the rst equation by 7 wL7 the second by 7R and add Similar for Ag We nd if L V R E wL w 4 A1 w 220 A w 220 31 CW 7mw iFo b 27 k 2 2 3 A2 UZTw Z WSW t For spring7 A1 t 7 E ii iwL t 7R 39 t L11 7 w ZQQU cosw ZQQU s1nw 7 V0 i 7 wL t R t 7 wZltwgt ZW cosw ZW s1nw cosw gimp qpt coswt 7 go 1 Where Zw lt7 7 wLgt 2 R2 cosgo L7 singo w we W So after a long time eg When 7 gtgt 1 such that go t 7gt 0 7 V0 t7gtoo V0 qt 7 5192 fazed coswt 7 so 7gt iwzw coswt 7 go rawSign t t steadystate 7 d4 V0 2t 7 a 7gt 7Zltwgt s1nwt 7 go cosz 7 sinz 7 cos wt 7g0 call this the steady state response Check cosz cos g 7 sinzsing 0 7 sinzi So the long time 7 gtgt 1 output of an LRC circuit With applied voltage V0 coswt is a current ipt With Vo amplitude ZOO and Which follows the cosine oscillation With a phase lag of go 7 I steady state fresponse Mt 005W 80 i g Vt V0 coswt tw7 It is natural to ask For What values of w L7 C7 R is the ampli cation Th largest 1 7 1 2 7 L Z w 7 W 1s clearly max1mum When L7 C and w are chosen such that w 7 Lci As R decreases response gets bigger Note As w 7gt 07 7 0 As w 7gt 007 7 0 when if we say the circuit is tuned When 0J2 cosgo 07 singo 17 go g and the phase lag of the output current is zero V t ipt LC coswt Y The current follows the time dependent voltage as if Ohmls Law were valid On radios7 usually the tunable parameter is C res onse p Let be frequencies of 77stations77i If you want to listen to M4 say7 tune C such that w4i Then the response V1 7r V4 7 7 t 7 7 7 t 0 2270 ZM cosltw1 sm 2 ZW cosw4 The ratio is very small Therefore you hear mostly The current causes a diaphragm to oscillate Z004 W1th the slgnal carr1ed by V4 Wthh varies slowly compound to the slgnal perlod i and Which causes the music signal the music on station 4 Likewise for a forced7 damped spring mt cosw 7 so Vpt cos Lute goi i 7 Where W f gt2 va W v sinso 7 k The largest velocity response occurs When w k m are chosen such that 0J2 E namely When the forcing frequency equals the natural frequency of the undamped spring we call this resonancei Forw2g0 Zbancl vpt coswt b Note in that case7 the function term exactly balances the force and the acceleration and spring force are equal and opposite equal and opposite d2 d m d 12 kzp Focoswt equal
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