Question

The differential equation describing the motion of a mass attached to a spring is x'' + 16x = 0. If the mass is released at t = 0 from 1 meter above the equilibrium position with a downward velocity of 3 m/s, the amplitude of vibrations is

Please show all work. Thank you!

Answer 1

Given differential equation x"' +16 x=0 Characteristic equation is p²+16=0 D's -16 -) D = 441 compler roots & Solution is X(t)= C, Cos (et) + C₂ Sin At) Given that mass released at t=0 from I meten ) X10)= 1 = C, Cos (4(0)). +62 Sin(460)) => ) = Ci Cosco) + (z Sinfo) ci (1( 20) = T 3 1 136 11 CIE!! and given that down word velocity = 3 m/s x'(0) = 3 x'(t)= = (* (t)) = -4C Sin (44) + 4G COS(44) X(0)=3 = 3= -46, Sin(o) + 462 Cos (0) 93=462 ) (2 894 Amplitute of vibrations A= JQ²+2 = V12+ (3) 1679 ale camscan voeten w = 1,25 Amplitude CSScanned with CamScanner