Actual Question
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)=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 equation
Answer:
A = 7.2/2329 and B = 0.75/2329
y = 7.2/2329 cos(t) + 0.75/2329 sin(t)
Step-by-step explanation:
Given
Mass, m = 2kg
Stretch/Displacement, d = 20cm
d = 0.2m
Friction Constant, Fc = 10
y(0) = 0.5
Driving force function F(t)=0.3cos(t).
Calculating spring constant, k from
d=mg/k
k = mg/d
k = 2 * 9.8/0.2
k = 98
Given than F = ma where m = mass and a = acceleration
The 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 get
2a = -98y -10v + 0.3cos(y) --- in terms of y
2 d²y/dt = -98y -10y' + 0.3cos(y) --- divide through by 2
d²y/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 t
y = Acos(t) + Bsin(t)
Differentiate
y'' = -Asin(t) + Bcos(t)
Different further
y'' = -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 comparison
48B - 5A = 0
48A + 5B = 0.15
From (48B - 5A = 0 )
5A = 48B
A = 48B/5
In (48A + 5B = 0.15)
48 * 48B/5 + 5B = 0.15
2304B/5 + 5B = 0.15
(2304B + 25B)/5 = 0.15
B = 5 * 0.15/2329
B = 0.75/2329
A = 48B/5
A = 48/5 * 0.75/2329
A = 7.2/2329
y = 7.2/2329 cos(t) + 0.75/2329 sin(t)
