Question

A mass weighing 16 lb stretches a spring 3 in. The mass is attached to a viscous damper with a damping constant of 2 lbs/ft. If the mass is set in motion from its equilibrium position with a downward velocity of 6 in/s, find its position u at any time t. (Use g = 32 ft/s2 for the acceleration due to gravity. Let u(t), measured positive downward, denote the displacement in feet of the mass from its equilibrium position at time t seconds.)

I have gotten 1.5927 but says it's wrong. Thanks in advance

Answer 1

Page Solution: Ilb= 0:45359237 kg 16 lb = 1680.45359237 1 kg = 7.26 kg 1 inch 0.0254 meters 3 inch 0.0762 meters. 6 inch = 0.1526 meters 1 ft = 0.305 meters go 29 ft 152 | 32X0:305 ) om 152 9.76 mis² ky: due to in the mass kt Now Intially lo be its natural lenger say yi be the llongation 7 7. 26 Kg spring attached constant mg spreng Equilibrium condition Kyi mg (7026)1976) Yi= 0.0762m kyiz k= (7.26)19.76) = 930 Nm 0.0762 R = 930 N/m.

= ab-g ft С= а( б•4535931)(416) -S 005 m . - 39. 03 N-s | e

Date Page Now Framing - at ky visons fund at t=o Tacion w Girls liguid) mg at accreation (say) Nelton e second law mass x accreation -ky-on ma fnet mat be try = 0 is dy + ky dt mary at 2 + =o day to dyt ky zo at² m at rewriting dzy te dy + kyo at2 m at m day + 29:03 dy 't 930 y zo 7026 at •7•26 atz dy + 3,998 dy t 128.09 at2 at an for somplicely lake it 8.998 = 4 (28.09 = 128

+ 20 equation became d²y Ady + 128 y por Auxiliary so lutuin let diy N 12 dy-t, V at2 at 10 عل dt? at 22 + 4x + 128 = 0 q= -4+ J 16- 4(1)(128) g a = - 4+ 116-512 q d = - 4 1 22.27 i 2 d=-gt 11.14 i 7 =-2-11. 41 General solution for of d- atbi gt) = et at [ A cor bt + A sin bt] a = -2; b= 11.14 que? A cos[!1+144] +8 siak[411 we can find this solution conditione -2t Now is two

Page here ylt) ulto) Released from equitienim position) A (1) + BLO) = 0 e (A)=0 A=0 dulti re 1.524 mis at t=0 2tr Asin (11:14+) (114) + Bucs (11.14911.14 1.524m's e - Blll. 14) = 1.524 B 0.1368 -2t 011368 cos (II:14€2] е so ult) e-2t cos (11-14+) ult) = 0.1368 e