Question

please find amplitude and freq of the steady state solution

An 8-kg mass is attached to a spring hanging from the ceiling and allowed to come to rest. Assume that the spring constant is 20 N/m and the damping constant is 2 N-sec/m. At time t= 0, an external force of 4 sin 2t cos 2t is applied to the system. Determine the amplitude and frequency of the steady-state solution.

Answer 1

a kg spoing hanging ceiling come damping attached is to Question: An 8 mass from the and to rest. Assume that the allowed to is 20 N/m. and the da constant At time 2 N-see/m. to, an constant is external is force 4 sin (2+) COS (2+) applied to the Determine system. the amplitude and -state steady solution spoing of frequency of the solution: Jale assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion. Thus, こ c du damping force dt where, is constant С called positive damping Constant. second Law : Thus, from Newton's

m du force t force damping = Restoring dt² + External force. d²n -ku- cda + f(t) m 11 다. dt dt² du m c da eq?(1) 1 + ka= f(t) + dt dt Equation of Motions, This the we K= 20 N/m have that: m=8kg, c=2 N-see/m, & fit) = 4 sin (at) cos(at) above values in equation (1): substituting 8 du dt² 2 da + 4 sin (2t) cos(at) 200= + dt 4 du dt2 da dt + loa= + 2 sin (at) cos(at) { 2 sin A Cosa= sin(2A) 07, du sin (44) 4. dt² + du dt + 10 x = differential order second This is Equation.

is also called The P.I. solution solution of DE state solution. steady let d at ED. P.I. Siin (4+) - (40² to +10) #PI Rules 22 4. P.I. sin (4t) 41-4²) +0+10 P.1. = sin (44) (D - 54) P.I. - (D+54) sin(4+) 2 D² - 54 P.1. - 4 Cos (4+) + 54 sin (4) (- 4²_54²) P.1. 1 (4 cos (4t) + 54 sin (4+)) 2932 state the Required This is steady solution

# Amplitude Amplitude is given by: Amplitude = 5 4² +54² 2932 (2932 * Amplitude - 2932 1 :. Amplitude m 2932 and 451 Angular frequency