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 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)|=

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

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)

in problems involving proportions, we can use cross products to test whether two ratios are equal and form a proportion. to find the cross products of a proportion, we multiply the outer terms, called the extremes, and the middle terms, called the means.

hope it