# A 2kg mass is attached to a spring hanging from the ceiling. This causes the spring to stretch 20 cm. The system has friction constant of 10. After coming to a stop at its new equilibrium, the mass is pulled 50 cm further toward the floor (i.e. y(0) and released subject to a driving force function F(t)=.3cos(t). Recall that we can calculate k when we know the displacement d caused by a mass m because d=mg/k [use g=9.8 m/sec2]. The differential equation to solve has k=98 1. The steady-state so

###### Question:

Recall that we can calculate k when we know the displacement d caused by a mass m because d=mg/k [use g=9.8 m/sec2].

The differential equation to solve has k=98

1. The steady-state solution yp =

2. The actual solution y(t) to the ivp is =

3. To illustrate that yp is steady-state and yh is transitory,

Calculate the value |yp(3)?y(3)|=

## Answers

Actual QuestionA 2kg mass is attached to a spring hanging from the ceiling. This causes the spring to stretch 20 cm. The system has friction constant of 10. After coming to a stop at its new equilibrium, the mass is pulled 50 cm further toward the floor (i.e. y(0) and released subject to a driving force function F(t)=0.3cos(t).Recall that we can calculate k when we know the displacement d caused by a mass m because d=mg/k [use g=9.8 m/s²]. Model this situation as a differential equationAnswer:A = 7.2/2329 and B = 0.75/2329y = 7.2/2329 cos(t) + 0.75/2329 sin(t)Step-by-step explanation:GivenMass, m = 2kgStretch/Displacement, d = 20cmd = 0.2mFriction Constant, Fc = 10y(0) = 0.5Driving force function F(t)=0.3cos(t).Calculating spring constant, k from d=mg/kk = mg/dk = 2 * 9.8/0.2k = 98Given than F = ma where m = mass and a = accelerationThe force governing the spring is;-ky - (Fc)v + F(t)F = ma = -ky - (Fc)v + F(t)ma = -ky - (Fc)v + F(t)By substitution, we get2a = -98y -10v + 0.3cos(y) --- in terms of y2 d²y/dt = -98y -10y' + 0.3cos(y) --- divide through by 2d²y/dt = -49y - 5y' + 0.15cos(y)d²y/dt can be written as y''y'' = -49y - 5y' + 0.15cos(y)y'' + 5y' + 49y = 0.15cos(y)Given the form that y is a function of ty = Acos(t) + Bsin(t)Differentiatey'' = -Asin(t) + Bcos(t)Different furthery'' = -Acos(t) - Asin(t)By substitution, y'' = 49y - 5y' + 0.15cos(y) becomes-Acos(t) - Asin(t) + 5(-Asin(t) + Bcos(t)) + 49(Acos(t) + Bsin(t)) = 0.15cos(t)-Acos(t) - Asin(t) -5Asin(t) + 5Bcos(t) + 49Acos(t) + 49Bsin(t) = 0.15cos(t)48Acos(t) + 48Bsin(t) -5Asin(t) + 5Bcos(t) = 0.15cos(t)(48A + 5B) cos(t) + (48A - 5B) sin(t) = 0.15cos(t)By comparison48B - 5A = 0 48A + 5B = 0.15From (48B - 5A = 0 )5A = 48BA = 48B/5In (48A + 5B = 0.15)48 * 48B/5 + 5B = 0.152304B/5 + 5B = 0.15(2304B + 25B)/5 = 0.15B = 5 * 0.15/2329B = 0.75/2329A = 48B/5A = 48/5 * 0.75/2329A = 7.2/2329y = 7.2/2329 cos(t) + 0.75/2329 sin(t)